Elementary Algebra Exercise Book I

ELEMENTARY ALGEBRA EXERCISE BOOK I inequAlities



Proof: If one of x,y,z is zero, without loss of generality, assume x =0, then |y + z| =0,
thus y + z =0, which implies x + y + z =0. If x,y,z are all nonzero, then there are four
possibilities:


1) If x,y,z are all positive, then y + z ≤ x, z + x ≤ y, x + y ≤ z , impossible.
2) If x,y,z have two positive one negative, without loss of generality, assume
x> 0,y > 0,z < 0. |y + z|≤ x ⇒−x ≤ y + z ≤ x ⇒ x + y + z ≥ 0. On the
other hand |x + y|≤ |z|⇒ x + y ≤−z ⇒ x + y + z ≤ 0. As a conclusion,
x + y + z =0.
3) If x,y,z have one positive two negative, without loss of generality, assume
x> 0,y < 0,z < 0 . |y + z|≤ x ⇒ y + z ≥−x ⇒ x + y + z ≥ 0. On the other
hand, |x + y|≤ |z|⇒ x + y ≤−z ⇒ x + y + z ≤ 0. As a conclusion,
x + y + z =0.
4) If x,y,z are all negative, then x ≤ y + z ≤−x, y ≤ z + x ≤−y, z ≤ x + y ≤−z ,
then x + y + z ≤ 2(x + y + z), thus x + y + z ≥ 0, a contradiction to the assumed
negativity condition.




3.41 If x> y > 0, show x − y + 2xy − y >x .
2
2
2
2 2 2 2 2 2 2 2
2
2
2
2
2
2
2
2
Proof 1: x>y > 0 ⇒ xy > y , 2xy − y >y ⇒ x − y >x − 2xy + y =(x − y) ⇒
x>y > 0 ⇒ xy > y , 2xy − y >y ⇒ x − y >x − 2xy + y =(x − y) ⇒

2
2
x − y >x − y x − y + 2xy − y >x − y + y = x .
2
2
2
2
2
2
x − y >x − y , and 2xy − y >y , thus

2 2 2 x − y >x − y
2
2
Proof 2: x> y > 0 ⇒ y> −y ⇒ x + y >x − y ⇒ x − y > (x − y) ⇒

2 2 2 2xy − y >y (ii).
2
(i). 2xy > 2y ⇒ 2xy − y >y ⇒

2
2
2
(i)+(ii)⇒ x − y + 2xy − y >x .

2
2
2
2
2
2

2
2
2

2
Proof 3: x − y + 2xy − y >x ⇔ x −y +2 2 (x − y )(2xy − y ) +2xy −y >x ⇔
2
2
2
2
2
2
2
2
2
x − y +
2xy − y >x ⇔ x −y +2
(x − y )(2xy − y ) +2xy −y >x ⇔
(x − y )(2xy − y ) >y − xy . The left hand side is greater than zero, while he right
2 2 2 2
(x − y )(2xy − y ) >y − xy
2
2
2
2

hand side y − xy = y(y − x) < 0, thus (x − y )(2xy − y ) >y − xy always holds.
2
2
2
2
2
1
3.42 Given x> 0,y > 0, + 9 y =1, show x + y ≥ 12.
x

Proof: Since x> 0,y > 0, we have 1 x + 9 y ≥ 2 1 x · 9 y = √ 6 xy . Since 1 x + y 9 =1 , we have
6 √ √
√ ≤ 1 , which is equivalent to xy ≥ 6. x + y ≥ 2 xy ≥ 12.
xy
3.43 a, b, c are real numbers and a + b + c =1, show a + b + c ≥ .
2
2
2
1
3
1
1
1
2
2
2
2
2
2
2
2
2
2
2
Proof 1: a + b + c =1 ⇒ c =1 − a − b , then a +b +c − = a +b +(1−a−b) − = a +b +1+a +b −2a−2b+2ab− = 2(a +
1
2
2
2
2
2
2
2
1
2

2
2
1
2
a +b +c 3 − = a +b +(1−a−b) − = a +b +1+a +b −2a−2b+2ab− = 2(a +
3
3
3
2 1 3 2 b−1 2 3 b−1 2 2 1 b−1 2 (3b−1) 2 2 ] ≥ 0
) + (3b−1)
1
1
1
2
2
2
1
2
2
2
2
2
2
2
1
2
2
2
) +b −b+ ] = 2[(a+ b−1 2
) −( b−1 2
b +ab−a−b+ ) = 2[a +(b−1)a+( b−1 2
2
2
1
2
2
2
2
2
2
2
2
2
2
1
a +b +c − = a +b +(1−a−b) − = a +b +1+a +b −2a−2b+2ab− = 2(a + b +ab−a−b+ 3 ) = 2[a +(b−1)a+( 2 ) −( 2 ) +b −b+ ] = 2[(a+ 2 ) + 12 12 ] ≥ 0
1

3
a +b +c
3 − = a +b +(1−a−b) − = a +b +1+a +b −2a−2b+2ab− = 2(a +
3
3
2
3
2
3
2
3
3
3
2
b−1 2
2
1
2
) +
(3b−1)
) +b −b+ ] = 2[(a+ b−1 2
2
b +ab−a−b+ ) = 2[a +(b−1)a+( b−1 2 b−1 2 2 2 1 3 1 3 b−1 2 (3b−1) 2 ] ≥ 0 .
2
1
) −( b−1 2
2 ) +
2 ) +b −b+ ] = 2[(a+
] ≥ 0
3 ) = 2[a +(b−1)a+(
2 ) −(
b +ab−a−b+
12
12
2
2
3
2
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101

ELEMENTARY ALGEBRA EXERCISE BOOK I inequAlities



2
Proof 2: a + b + c =1 ⇒ (a + b + c) =1 ⇒ a + b + c =1 − 2(ab + bc + ca) (i).
2
2
2
2
2

2
2
2
2
2
2
2
a + b ≥ 2ab, b + c ≥ 2bc, c + a ≥ 2ca , add them up to obtain 2(a + b + c ) ≥ 2(ab + bc + ca)
2(a + b + c ) ≥ 2(ab + bc + ca)(ii).
2
2
2
(i)+(ii)⇒ 3(a + b + c ) ≥ 1 ⇒ a + b + c ≥ .
2
2
2
2
2
1
2
3
3.44 The function f(x) is defined on [0, 1], and f(0) = f(1) . For any distinct
x 1 ,x 2 ∈ [0, 1], we have |f(x 2 ) − f(x 1 )| < |x 2 − x 1 | . Show |f(x 2 ) − f(x 1 )| < .
1
2
Proof: Let 0 ≤ x 1 <x 2 ≤ 1. We consider two cases:
1) If x 2 − x 1 ≤ , then |f(x 2 ) − f(x 1 )| < |x 2 − x 1 |≤ .
1
1
2 2
2) If x 2 − x 1 > , then from f(0) = f(1) we obtain
1
2
|f(x 2 ) − f(x 1 )| = |f(x 2 ) − f(1) + f(0) − f(x 1 )|≤|f(x 2 ) − f(1)| + |f(0) − f(x 1 )|
|f(x 2 ) − f(x 1 )| = |f(x 2 ) − f(1) + f(0) − f(x 1 )|≤|f(x 2 ) − f(1)| + |f(0) − f(x 1 )|. Hence, (1 − x 2 ) + (x 1 − 0) = 1 − (x 2 − x 1 ) < 1 2

(1 − x 2 ) + (x 1 − 0) = 1 − (x 2 − x 1 ) < .
1
2
3.45 The equation |x| = ax +1 has one negative root but no positive root, find the
range of the parameter a .
Solution 1: Let x be the negative root of the equation |x| = ax +1 , then
−x = ax +1 ⇒ x = −1 < 0, thus a +1 > 0. Equivalently, when a> −1, the equation has
a+1
a negative root.
1
Suppose the equation has a positive root x, then x = ax +1 ⇒ x = 1−a > 0 , thus a< 1.
As a conclusion, the condition that the equation has one negative root but no positive root
is equivalent to a> −1 holds but a< 1 fails, that is, a ≥ 1.

Solution 2: Another approach is to plot the functions y = |x| and y = ax +1 on the
Cartesian plane. This will directly give us the same conclusion.


2
3.46 Solve the inequality 3x −4x−23 > 2.
2
x −9
2
2
Solution: 3x −4x−23 > 2 ⇔ x −4x−5 > 0 ⇔ (x+1)(x−5) > 0. This inequality is equivalent to
2
2
x −9 x −9 (x+3)(x−3)
(x + 3)(x + 1)(x − 3)(x − 5) > 0 whose solution set is (−∞, −3) ∪ (−1, 3) ∪ (5, +∞).












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102

ELEMENTARY ALGEBRA EXERCISE BOOK I inequAlities




3.47 Consider the inequality x +2 >m(x − 1), (1) if the inequality holds for any real
2
number x , find the range of m ; (2) if for any m ∈ [−2, 2] the inequality holds, find the

range of x .


Solution: (1) x +2 >m(x − 1) ⇔ m(x − 1) − (x + 2) < 0 ⇔ mx − x − m − 2 < 0. This
2
2
2

m< 0
m< 0
m< 0

inequality holds for any real number x , then m< 0 ⇔ m< 0 ⇔ m< √0 √ ⇔
2
Δ = 1 +4m(m + 2) < 0 ⇔ 4m +8m +1 < 0 ⇔ −1 − 3 √ <m< −1+ 3 √ 3 ⇔
2
3
√ Δ = 1 +4m(m + 2) < 0 4m +8m +1 < 0 −1 2 − 2 <m< −1+ 2 2
√
m< 0 m< 0 m< 0 3 √ 3 √
3
−1 −
<m < −1+
⇔ ⇔ √ √ ⇔ −1 2 − <m < −1+ 2 3 .
2
Δ = 1 +4m(m + 2) < 0 4m +8m +1 < 0 −1 − 3 <m< −1+ 3 2 2
√ √ 2 2
3 3
2
−1 − <m < −1+ (2) For any m ∈ [−2, 2] the inequality holds. Let f(m) =(x − 1)m − (x + 2) which is a
2 2
linear function of m , and f(m) < 0 should hold for any m ∈ [−2, 2], equivalently we
2 2 2 2 1 1
−2(x − 1) − (x + 2) < 0
2x + x> 0
should have −2(x − 1) − (x + 2) < 0 ⇔ 2x 2 2x + x> 0 ⇔ x> x> 0 or x< − 2 √ 1+ 33 ⇔ ⇔
x> 0 or x< −

1
2
−2(x − 1) − (x + 2) < 0
⇔+ x> 0
2 √
√
⇔ 0 or x< −
√
2
2
1− 33
2
2x − x − 4 < 0
2
2(x − 1) − (x + 2) < 0
1− 33
1+ 33
2 √
√
2(x − 1) − (x + 2) < 0 ⇔
2x − x − 4 < 0 ⇔
2
2
< x< 33
1+
2(x − 1) √ − √ (x + 2) < 0 2x − x − 4 < 0 1− 33 4 < x< < x< ⇔ 4
4
4
0 < x< 1+ 33 √ 4 4
1+ 33
√
0 < x< 33
1
0 < x< 1+ 4 or 4 1− 33 < x< − .
4 4 2
3.48 x,y,z are positive numbers, and xyz(x + y + z) =1. Find the minimum value
of (x + y)(x + z).
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1 10303

ELEMENTARY ALGEBRA EXERCISE BOOK I inequAlities




Solution: The given conditions imply that (x + y)(x + z) = yz + x(x + y + z) ≥ 2 yz · x(x + y + z)=2
√

(x + y)(x + z) = yz + x(x + y + z) ≥ 2 yz · x(x + y + z)=2. When x = 2 − 1,y = z =1, the equal sign is reached in the
above inequality, thus the minimum value of (x + y)(x + z) is 2.

2
3.49 Real numbers a, b, c satisfy a + b + c =1. Show that one of |a − b|, |b − c|, |c − a|
2
2
√
is not greater than 2 .
2
Proof: Without loss of generality, we assume a ≤ b ≤ c , and let m be the minimum one

of |a − b|, |b − c|, |c − a| . Then b − a ≥ m, c − b ≥ m, c − a =(c − b) + (b − a) ≥ 2m . On

2 2 2 2 2 2 2 2 2

one hand, (a − b) + (b − c) +(c − a) = 2(a + b + c ) − 2(ab + bc + ca)=3(a + b + c ) −
2
2
2
2
2
2
2
2
2
(a − b) + (b 2 − c) +(c − a) = 2(a + b + c ) − 2(ab + bc + ca)=3(a + b + c ) −
(a + b + c) ≤ 3
2
2
2
2
2
2
(a + b + c) ≤ 3. On the other hand, (a − b) +(b − c) +(c − a) ≥ m + m + (2m) =6m 2
2
√
2 2 2 2 2 2 2 2 2
(a − b) +(b − c) +(c − a) ≥ m + m + (2m) =6m . Hence, 6m ≤ 3 ⇒ m ≤ .
2
α β α+1 β+1
3.50 Let α, β be real numbers, show log (2 +2 ) ≥ 2 + 2 .
2
α β α+1 β+1 α β α+β+2
Proof: To show log (2 +2 ) ≥ 2 + 2 , we only need to show 2 +2 ≥ 2 2 , we only need
2
to show 2 2α +2 α+β+1 +2 2β ≥ 2 α+β+2 , we only need to show 2 α−β−2 +2 −1 +2 β−α−2 ≥ 1, we
1
2
1
only need to show 2 α−β +2+ α−β ≥ 4, we only need to show (2 α−β α−β ) ≥ 0 which is
2 −
2
2 2
obviously valid.
2
3.51 Given the function f(x) = ax − c that satisfies −4 ≤ f(1) ≤−1, −1 ≤ f(2) ≤ 5.
Find the range of f(3).
1
1
2
Solution: From f(x) = ax − c, we know f(1) = a − c, f(2) = 4a − c , thus a = [f(2) − f(1)],c = [f(2) − 4f(1)]
3 3
1
5
1
8
1
a = [f(2) − f(1)],c = [f(2) − 4f(1)] . Then f(3) = 9a − c = 3[f(2) − f(1)] − [f(2) − 4f(1)] = f(2) − f(1)
3 3 3 3 3
1 8 5 8 5 8 5
f(3) = 9a − c = 3[f(2) − f(1)] − [f(2) − 4f(1)] = f(2) − f(1). Hence, × (−1) + (− ) × (−1) ≤ f(3) ≤ × 5+(− ) × (−4), that is, −1 ≤ f(3) ≤ 20
3 3 3 3 3 3 3
−1 ≤ f(3) ≤ 20.
1
1
3.52 Given real numbers a> 0,b > 0,c > 0 and a + b + c =1, show + + 1 ≥ 9.
a b c
√
Proof 1: The conditions together with Cauchy’s Inequality imply a+b+c ≥ 3 abc ⇒ √ 1 ≥ 3 =3 .
3 3 abc a+b+c
1
1 + + 1
1
Apply Cauchy’s Inequality again to obtain a b c ≥ 3 1 · · 1 = √ 1 ≥ 3 ⇒ 1 + 1 + 1 ≥ 9 .
3 a b c 3 abc a b c
2
2
2
2
2
2
Proof 2: 1 1 + 1 bc+ac+ab − 9= (a+b+c)(bc+ac+ab)−9abc 2 a c+a b+b c+ab +ac +bc −6abc =
2
2
2
2
2
1 + 1 a + + 1 b − 9= − 9=+ab abc − 9= (a+b+c)(bc+ac+ab)−9abc = a = c+a b+b c+ab +ac +bc −6abc =
bc+ac
c
abc
abc
a a(b−c) +b(c−a) +c(a−b) 2 abc abc
2 c
abc
b
2
2
2
a(b−c) +b(c−a) +c(a−b) 2 ≥ 0 . ≥ 0
abc
abc
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104

ELEMENTARY ALGEBRA EXERCISE BOOK I inequAlities




1 2 1 2 25
3.53 Let a> 0,b > 0 and a + b =1, show (a + ) +(b + ) ≥ .
a b 2
√ √
1 1 1 ≥ 4.
Proof 1: 1= a + b ≥ 2 ab ⇒ ab ≤ ⇒ ab ≤ ⇒
2 4 ab
And (a+ ) +(b+ ) ) ≥ [ ≥ [ a+ +b+ ] = (1 ++ 1 1 + ) = (1 ++ 1 2 2 2525,
1 1
1 2 1 2
1 2 1 2
1 1
a+ +b+
(a+ ) +(b+
b 2 b 2
1 1
1 2 1 2
1
1 1
+ ) = (1
)
]
a a
b b
a a
) ≥≥
2 2 2 2 4 = (1 a a b b 4 4 abab 4 4
4
1 2 1 2 25
thus (a + ) +(b + ) ≥ .
a b 2
2
1
2
2
2
2
1 2
2
2
2
2
1 2
2
Proof 2: Let a = sin α, b = cos α , then (a+ ) +(b+ ) = (sin α+csc α) +(cos α+sec α) ≥ (sin α+csc α+cos α+
2
a
b
2
2
1
2
2
2
1
2
4
1
1
1
1
sec α) = (1+ sin α cos α 2 1 2 sin α cos α ) = (1+ sin 2α ) = (1+4 csc 2α) ≥
2 +
2 ) = (1+
2
2
2
2
2
2
2
1 11
2
2 2
2 2
2 2
2 2
22
1 2 1 2
2 2
2
2 2
2 25 2
1 2 1 2
(a+ ) +(b+ ) = (sin α+csc α) +(cos α+sec α) ≥ (sin α+csc α+cos α+
(1 + 4) =
(a+ ) +(b+ ) = (sin α+csc α) +(cos α+sec α) ≥ (sin α+csc α+cos α+
2
2
2 2
2
2
1 2
2 2 2 1
1
2
1 2
2
2 2
2 2
2
2
2
b1 2
2 ) +(b+ ) = (sin α+csc
2
2
2
a 1 2
(a+
2
b
2 2 (a+ ) +(b+ ) = (sin α+csc α) +(cos α+sec α) ≥ (sin α+csc α+cos α+
a
1 1
4 4
2 2
1
2 2
2 2
1
1
1 12
2 2
1 2
1 1 2
1
1 1 2
2
2
2
2
2
2
) = (1+ +
2 ) =
a
sec α) =
(a+
) = (1
sec α) = (1+ (1+ b 1 b b 1 2 1 cos α α 1 2 2 2 1 1 α) +(cos α+sec α) ≥ (sin α+csc α+cos α+
) = (1+4 csc 2α) ≥α+
1 a ) +(b+ ) = (sin α+csc α) +(cos α+sec α) ≥ (sin α+csc α+cos
2
2 + +
2 ) = (1+ (1+
2
2
) = (1+4 csc 2α) ≥
2
2 2
1
2 2
2
2 2
2 2 1
1
sin α cos α α 2
cos 1
2
sin 4
1
2
2 21
1
2
2 1
sin 2α 2α4 2
2 sin α cos 1
2 a
2
1
sin α sin α
2 2 1
2
2
sec α) =
1 sec α) = (1+ 2 +
2 + 2 ) = (1+
2 (1+
) = (1+4 csc 2α) ≥
25
2
1
2
1
2
) = (1+ 2 4
2
1
2 2
2
1
1
(1 + 4) =
1
) =
2 +
sec α)
sin α cos α ) = (1+
2
1 (1 + 4) == (1+ sin α cos α 2 ) = (1+ 2 2 1 2 2 α ) = (1+ sin 2α ) = (1+4 csc 2α) ≥
cos α ) = (1+
2
sin α
2
2
sin 2α 2
sin α cos
25 2
2
2
2 (1+4 csc 2α) ≥
25
2
2
2
2
2 1
2
2
2 1 (1 + 4) = 2 2 2 25 sin α cos α 2 sin α cos α 2 sin 2α 2
2 .
2
2 1 (1 + 4) = 25
2
2 (1 + 4) =
2 2
3.54 Let a,b,c,d,m,n be positive real numbers,
√ √ √
d
P = ab + cd, Q = ma + nc b + . Compare P and Q .
m n
√
d
2
2
Solution: P = ab + cd +2 abcd, Q =(ma + nc)( b + )= ab + cd + nbc + mad . Since
m n m n
√
2
2
nbc + mad ≥ 2 nbc · mad =2 abcd , then P ≤ Q . Because P, Q are positive, we have P ≤ Q .
m n m n
3.55 Show the inequality (a + b) ≤ 128(a + b ).
8
8
8
Proof: (a − b) ≥ 0 ⇒ a + b ≥ 2ab. Similarly we have a + b ≥ 2a b ,a + b ≥ 2a b .
2
8
2 2
8
4
4
2
4 4
2
Add a + b ,a + b ,a + b to the above three inequalities respectively to obtain
8
2
4
2
8
4
2(a + b ) ≥ (a + b) , 2(a + b ) ≥ (a + b ) , 2(a + b ) ≥ (a + b ) ) . The last inequality leads
22
4 4
22
8 8
22
2
2 2 2
8 8
2 2
4
4 2 2
4 4
44
2(a + b ) ≥ (a + b) , 2(a + b ) ≥ (a + b ) , 2(a + b ) ≥ (a + b

2 2
2 2
8 8 8 8 4 4 4 2 4 2 4 4 4 4 2 2 22 2 2 2 2 2 2 2 2 2 2

128(a + b ) ≥ 64(a + b ) = 16[2(a + b )] ≥ 16[(a + b ) ] = {[2(a + b )] } ≥
to 128(a + b ) ≥ 64(a + b ) = 16[2(a + b )] ≥ 16[(a + b ) ] = {[2(a + b )] } ≥
2
2 2
2
4
2 2 2
4
2
4
4 2
8
8
2
128(a + b
2 2 2 2 2 2 ) ≥ 64(a + b ) = 16[2(a + b )] ≥ 16[(a + b ) ] = {[2(a + b )] } ≥
{[(a + b) ] } =(a
8
{[(a + b) ] } =(a + b) + b) 8
{[(a + b) ] } =(a + b) .
2 2 2
8
2
3.56 Given the function f(x − 3) = log x 2 2 (a> 0,a =1 ) that satisfies
a 6−x
f(x) ≥ log 2x . Find the domain of the function f(x) .
a
3+t
Solution: Let x − 3= t , then x = 3+ t . Substitute it into the function: f(t) = log a 3−t , thus
2
2
f(x) = log 3+x . Then the inequality f(x) ≥ log 2x is equivalent to log 3+x ≥ log 2x .
a 3−x a a 3−x a
⎧
3+x > 0
⎨ 3−x
If a> 1, then 3+x ≥ 2x ⇒ x ∈ (0, 1) ∪ [− , 3).
3
3−x 2
⎩
x> 0
⎧
3+x >
< 0
⎨ 3−x
If 0 <a < 1, then 3+x ≤ 2x ⇒ x ∈ [1, ) .
3
3−x 2
⎩
x> 0
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105

ELEMENTARY ALGEBRA EXERCISE BOOK I inequAlities




3.57 Given a< −1, and x satisfies x + ax ≤−x , and x + ax has the minimum
2
2
1
value − , find the value of a .
2
2 2 a 2 a 2
Solution: a< −1,x + ax ≤−x ⇒ x[x +(a + 1)] ≤ 0 ⇒ 0 ≤ x ≤−(a + 1). Let f(x) = x + ax =(x + ) − 4

2
2 a 2 a 2
f(x) = x + ax =(x + ) − .
2 4
If −(a + 1) < − ⇔−2 <a < −1, then f(x) reaches its minimum value f(−a − 1) = a +1
a
2
3
1
at x = −(a + 1), thus a +1 = − ⇒ a = − .
2 2
2
a
If −(a + 1) ≥− ⇔ a ≤−2, then f(x) reaches it minimum value − at x = − , thus
a
a
2 4 2
√
a 2 1
− = − ⇒ a = ± 2 both of which violate a ≤−2.
4 2
As a conclusion, a = − .
3
2















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106
106

ELEMENTARY ALGEBRA EXERCISE BOOK I inequAlities




3.58 a 1 ,a 2 , ··· ,a n are positive numbers and satisfy a 1 a 2 ··· a n =1 , show
(2 + a 1 )(2 + a 2 ) ··· (2 + a n ) ≥ 3 .
n
√
3
Proof: Use an arithmetic mean-geometric mean inequality a + b + c ≥ 3 abc (a, b, c are
√
positive numbers) to obtain 2+ a i = 1+ 1+ a i ≥ 3 a i (i =1, 2, ··· ,n ). Then
3
√√
n n 3 3 n n

(2 + a 1 )(2 + a 2 ) ··· (2 + a n ) ≥ 3 ·(2 + a 1 )(2 + a 2 ) ··· (2 + a n ) ≥ 3 · a 1 a 2 ··· a n =3a 1 a 2 ··· a n =3 .
3.59 If a, b, c are side lengths of a triangle, show a b(a − b)+ b c(b − c)+ c a(c − a) ≥ 0
2
2
2
and determine when the equal sign is reached.
Proof: Let a = y + z, b = z + x, c = x + y where x,y,z are positive numbers. Substitute

them into the inequality:
22 22 22
(y + z) (x + z)(y − x)+(z + x) (x + y)(z − y)+ (x + y) (y + z)(x − z) ≥ 0 ⇔(y + z) (x + z)(y − x)+(z + x) (x + y)(z − y)+ (x + y) (y + z)(x − z) ≥ 0 ⇔
2
2
2
(y + z) (x + z)(y − x)+(z + x) (x + y)(z − y)+ (x + y) (y + z)(x
− z) ≥ 0 ⇔
33
33
33
x
x z + y x + z y − xyz(x + y + z) ≥ 0z + y x + z y − xyz(x + y + z) ≥ 0
x z + y x + z y − xyz(x + y + z) ≥ 0. Divide both sides by xyz to obtain
3
3
3
x 2 + y 2 + z 2 ≥ x + y + z which can be proven by the inequalities
y z x
2 2
2 2
2 2
xx + y ≥ 2x,+ y ≥ 2x, yy + z ≥ 2y,+ z ≥ 2y, zz + x ≥ 2z+ x ≥ 2z .
yy zz xx
These inequalities have the equal sign if and only if x = y = z , that is, the original inequality
has the equal sign if and only if a = b = c .
|a+b| |a| + |b|
≤ 1+|b| .
3.60 a, b are real numbers, show 1+|a+b| 1+|a|
=
=
|a+b| 1+|a+b|−1 1 1 |a|+|b| |a| |b|
+
1
1
|a|
|a|+|b|
|b|
1+|a+b| =
Proof: Since |a + b| ≤|a| + |b|, we have |a+b| = 1+|a+b|−1 =1 − 1+|a+b| ≤ 1 − 1+|a|+|b| = 1+|a|+|b| = 1+|a|+|b| + 1+|a|+|b| ≤
≤
≤ 1 −
1+|a+b| =1 −
1+|a+b| 1+|a+b| 1+|a+b| 1+|a|+|b| 1+|a|+|b| 1+|a|+|b| 1+|a|+|b|
|a+b| 1+|a+b|−1 1 1 |a|+|b| |a| |b| |a| + |b|
= =1 − ≤ 1 − = = + ≤ 1+|a| + 1+|b|.
|a|
|b|
1+|a+b| 1+|a+b| 1+|a+b| 1+|a|+|b| 1+|a|+|b| 1+|a|+|b| 1+|a|+|b| 1+|a| 1+|b|
|a| |b|
+
1+|a| 1+|b|
1
y
x
2
3.61 Given 0 <a < 1,x + y =0, show log (a + a ) ≤ log 2+ .
a
a
8
√ x+y
y
x+y
x
Proof: a + a ≥ 2 a a =2a 2 . Since 0 <a < 1, we have log (a + a ) ≤ log (2a x+y x+y x+y = log 2+ x−x 2 = log 2+ x(1 − x) ≤
x y
1
x
y
2 ) = log 2+
2
2
a
x
a x
y
y
2 ) = log 2+) = log 2+
log (a +(a + a ) ≤ log (2a 2 a x+y 2 x+y a x−x 2x−x a 1 2 1
= log 2+= log 2+
1 x+1−x 2 a ) ≤ log (2alog
= log 2+ x(1 − x) ≤= log 2+ x(1 − x) ≤
a
a
a
(
log a 2x+1−x 2 ) = log 2+ 1 a a a 2 2 a a 2 2 a a 2 2
x+y
) = log 2+) = log 2+ .
1
a
1
2
y
1
x
(
log
log (a + a ) ≤ log (2a 2 ) = log 2+ x+y = log 2+ x−x = log 2+ x(1 − x) ≤ log 2+ ( 2 x+1−x 2 a a 1 8 8 1

8
a
a
a
2
a 2 a 2
2
a
a
2
2
2
log 1 x+1−x 2 1
(
) = log 2+
a 2 2 a 8
3.62 The system of inequalities
√
2
x − 2x − 8 < 8 − x
2
x + ax + b< 0
has the solution 4 ≤ x< 5, find the conditions a and b should satisfy.
⎧ 2 ⎧
⎨ x − 2x − 8 ≥ 0 ⎨ x ≤−2 or x ≥ 4
Solution: √ 2 ⇒ x ≤−2
x − 2x − 8 < 8 − x ⇒ 8 − x> 0 ⇒ x< 8
⎩ 2 2 ⎩ 36
x − 2x − 8 < (8 − x) x<
7
36
or 4 ≤ x ≤ 7 .
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107

ELEMENTARY ALGEBRA EXERCISE BOOK I inequAlities




The solution of the inequality x + ax + b< 0 should have the form α <x < β . Since the
2
solution of the inequality system is 4 ≤ x< 5, then β =5, −2 ≤ α< 4. Since β =5, then

25 + 5a + b =0 Since α + β = −a , then 3 ≤−a< 9, that is, −9 <a ≤−3. As a conclusion,

a, b should satisfy −9 <a ≤−3
5a + b + 25 = 0.



2
2
2
2
3.63 If x,y,z ≥ 1, show (x − 2x + 2)(y − 2y + 2)(z − 2z + 2) ≤ (xyz) − 2xyz +2
(x − 2x + 2)(y − 2y + 2)(z − 2z + 2) ≤ (xyz) − 2xyz +2.
2
2
2
2
2 2 2 2 2 2
Proof: Since x ≥ 1,y ≥ 1, we have (x −2x+2)(y −2y+2)−[(xy) −2xy+2] = (−2y+2)x +(6y−2y −4)x+(2y −

2
2
2
2
2
2
(x −2x+2)(y −2y 2 +2)−[(xy) −2xy+2] = (−2y+2)x +(6y−2y −4)x+(2y −
2
2
4y +2) = −2(y −1)x −2(y −1)(y −2)+2(y −1) = −2(y −1)[x +(y −2)x+1−y]=
2
2
2
2
2
2
2
2 2y+2)−[(xy) −2xy+2] = (−2y+2)x +(6y−2y −4)x+(2y −
2
(x −2x+2)(y −
2 2 2 2 2 2 4y +2) = −2(y −1)x −2(y −1)(y −2)+2(y −1) = −2(y −1)[x +(y −2)x+1−y]=
(x −2x+2)(y −2y+2)−[(xy) −2xy+2] = (−2y+2)x +(6y−2y −4)x+(2y − −2(y − 1)(x − 1)(x + y − 1) ≤ 0 2 2
2
2
2
2
2
2
2
2 −2y+2)−[(xy) −2xy+2] =
(x −2x+2)(y
−2(y − 1)(x − 1)(x + y − 1) ≤ 0
2
2(−2y+2)x +(6y−2y −4)x+(2y −
4y +2) = −2(y −1)x −2(y −1)(y −2)+2(y −1) = −2(y −1)[x +(y −2)x+1−y]= 4y +2) = −2(y −1)x −2(y −1)(y −2)+2(y −1) = −2(y −1)[x +(y −2)x+1−y]=

2
2
2
2
4y +2) = −2(y −1)x −2(y −1)(y −2)+2(y −1) = −2(y −1)[x +(y −2)x+1−y]=
−2(y − 1)(x − 1)(x + y − 1) ≤ 0 2 2 −2(y − 1)(x − 1)(x + y − 1) ≤ 0, then (x − 2x + 2)(y − 2y + 2) ≤ (xy) − 2xy +2
−2(y − 1)(x − 1)(x + y − 1) ≤ 0

2
2
2
2
2
2
(x − 2x + 2)(y − 2y + 2) ≤ (xy) − 2xy +2(i). Similarly, since xy ≥ 1,z ≥ 1, we have [(xy) − 2xy + 2](z − 2z + 2) ≤ (xyz) − 2xyz +2
2
2
2
2
2
2
2
[(xy) − 2xy + 2](z − 2z + 2) ≤ (xyz) − 2xyz +2 (ii). From (i) and (ii), we can obtain (x − 2x + 2)(y − 2y + 2)(z − 2z + 2) ≤ (xyz) − 2xyz +2

(x − 2x + 2)(y − 2y + 2)(z − 2z + 2) ≤ (xyz) − 2xyz +2.
2
2
2
2
1
1
3.64 Given natural numbers a<b< c , m is an integer, and + + 1 = m , find
a b c
a, b, c .
1 1 1 5
Solution: Since a, b, c are natural numbers and a<b< c , we have a ≥ 1,b ≥ 2,c ≥ 3, 0 <m ≤ + + =1
1 2 3 6
1 1 1 5
a ≥ 1,b ≥ 2,c ≥ 3, 0 <m ≤ + + =1 . Since m is an integer, we have m =1 and a =1 . If a ≥ 3 , then
1 2 3 6
1
1
1
1 + + 1 1 + + 1 = 47 < 1. Hence, + + 1 = m =1. Therefore, a =2. Then + 1 1 = 1
1
1
a b c ≤ 3 4 5 60 a b c b c =1 − 2 2
1 + 1 1 = . If b ≥ 4, then + 1 1 + 1 = 9 1 1 1 1 = ,
1
1
1
b c =1 − 2 2 b c ≤ 4 5 20 < , thus b =3. Then =1 − − 3 6
c
2
2
thus c =6.
x
x
3.65 Given 1 <a < 2,x ≥ 1, and f(x) = a +a −x ,g(x)= 2 +2 −x . (1) Compare f(x)
2 2
m .
n 1 1
and g(x). (2) Let n ∈ N, n ≥ 1, show f(1) + f(2) + ··· + f(2n) < 4 − 2n
2 2
x
x
x x
x
x 2x
x
x
a −a
+1
1 a
Solution: (1)f(x)−g(x) = a +a −x − 2 +2 −x = ( 2x +1 2 2x x ) = 2 a +2 −2 2x x x = (a −2 )(2 a −1) .
2 2 2 a x − 2 2 x+1 x 2 x+1 x
a
a
x
x x
Since 1 <a < 2,x ≥ 1, then 2 a > 1,a < 2 , thus (a −2 )(2 a −1) < 0 , that is,
x
x x
x
x
a
2 x+1 x
f(x) − g(x) < 0. Hence, f(x) <g(x).
2n
1 1
1
2
(2) Since f(x) <g(x), then f(1) + f(2) + ···+ f(2n) <g(1) + g(2) + ···+ g(2n)= (2 + 2 + ···+2 )+ ( +
1 1
1
2n
2
f(1) + f(2) + ···+ f(2n) <g(1) + g(2) + ···+ g(2n)= 2 2 (2 + 2 + ···+2 )+ ( 2 2+
1
n 2 2
1
1
1
1
1
2n−1
n
2
2 + ··· +
2n ) = (1 +2 +2 + ··· +2
1
2n =4 − 1
2
n
)+ (1 − 1
1
1 1
2
2n ) < 4 − 1 +1 − 1
n
2n
f(1) + f(2) + ···+ f(2n) <g(1) + g(2) + ···+ g(2n)= (2 + 2 + ···+2 )+ ( + 1 2 + 2 ··· + 1 2n ) 2 = (1 +2 +2 + ··· +2 2n−1 )+ (1 2 − 2 ) < 4 − 1 +1 − 2 =4 − 2 2n
1
1 1
2n
2
f(1) + f(2) + ···+ f(2n) <g(1) + g(2) + ···+ g(2n)= 2 (2 + 2 + ···+2 )+ ( + 2 2 2 2 2n 2 2n 2 2n
2 2
1 1 2 2n−1 1 1 2 n 1 n 2 2 1
2 + ··· + 1
)+ (1 − 1
2n ) = (1 +2 +2 + ··· +2 2n−1
2
n
n
1 2 + ··· + 2 ) = (1 +2 +2 + ··· +2 )+ 1 2n ) < 4 − 1 +1 − 1 2n =4 − 1 2 .
2n
2 =4 −
2 ) < 4 − 1 +1 −
2 2 2 2n 2 2 (1 − 2 2n 2 2n 2 2n
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108

ELEMENTARY ALGEBRA EXERCISE BOOK I inequAlities




ab c


3.66 a, b, c are real numbers, and a + b + c< 0, show b ca ≥ 0.


ca b

2
2
2
2
Proof: a + b ≥ 2ab, b + c ≥ 2bc, c + a ≥ 2ca , and add them up to obtain
2
2
2 2
2 2
2(a + b + c ) ≥ 2(ab + bc + ca), thus (ab + bc + ca) − (a + b + c ) ≤ 0.
2 2
2(a + b + c ) ≥ 2(ab + bc + ca) 2 2 2

1

ab c
ab c a + b + cb c 1 bc bc 1 b 1 b c c







a + b + cb c

ab c a + b + cb c 1 bc 1 b c



1
b


1

ab
b ca =
=(a+b+c) 1
=
b ca = a + b + cca =(a+b+c) 1 ca ca =(a+b+c) 0 c − ba = − c c
=(a+b+c) 0 c − ba − c



c a + b + cca bc

a + b + cb c





=









b ca a + b + cca =(a+b+c) 1 ca =(a+b+c) 0 c − ba − c =
b







1 ab ab
a + b + ca b b
ca b a + b + ca =(a+ 1 0 a − bb − c bb − c =


ca bca a + b + cca b+c) 1 ca =(a+b+c) 0 c − ba − c


=



2 0 a − bb − c
ca b a + b + ca b 1 ab 0 a −

1 ab
2
(a + b + c)[−(b − c) − (a − b)(a − c)] = (a + b + c)[(ab + bc + ca) − (a + b + c )] ≥ 0 2 0 a − bb − c
2



a + b + ca b
ca b
2
2
2
(a + b + c)[−(b − c) − (a − b)(a − c)] = (a + b + c)[(ab + bc + ca) − (a + b + c )] ≥ 0
2


2 − (a − b)(a − c)] = (a + b + c)[(ab + bc + ca) − (a + b + c )] ≥ 0.
2
2
2
2
(a + b + c)[−(b − c)
2
2
2
(a + b + c)[−(b − c) − (a − b)(a − c)] = (a + b + c)[(ab + bc + ca) − (a + b + c )] ≥ 0
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109
109

ELEMENTARY ALGEBRA EXERCISE BOOK I inequAlities




2
x − x − 2 > 0
3.67 If the system of inequalities 2x +(5+2k)x +5k< 0 has only one integer
2
solution −2, find the range of k .


Solution: The solution of x − x − 2 > 0 is x< −1 or x> 2. The second inequality is
2
equivalent to (2x + 5)(x + k) < 0. When −k< − , i.e. k> , the second inequality has
5
5
2 2
5
5
the solution −k <x < − , in which −2 is not included. When −k> − , i.e. k< , the
5
2 2 2
second inequality has the solution − <x< −k , then the solution of the inequality system
5
2
x> 2
is x< −1 or − <x < −k . To have only one integer solution −2, we should
5
5
− <x < −k 2
2
5
have −k ≤ 3 and −k> −2, that is, −3 ≤ k< 2. When −k = − , i.e. k = , the second
5
2 2
inequality has no solution. As a conclusion, k ∈ [−3, 2).
3 .
a b c
3.68 Let a, b, c are positive numbers, show a b c ≥ (abc) a+b+c
3 , we only need
a b c
Proof: Without loss of generality, let a ≥ b ≥ c> 0. To show a b c ≥ (abc) a+b+c
b−c ≥ 1, we only need to show
3a 3b 3c
to show a b c ≥ (abc) a+b+c , we only need to show a a−b b b−a c c−a
a c−a b c−a c
a a−b b b−c c a−c a b a ≥ 1, thus the
a−c ≥ 1. Since a − b ≥ 0,b − c ≥ 0,a − c ≥ 0, we have ≥ 1, ≥ 1,
b a−b c b−c a b c c
last inequality holds.
2
2
2
2
2
2
3.69 If a,b,c,x,y,z are all real numbers, and a + b + c = 25,x + y + z = 36, ax + by + cz = 30

a+b+c
a + b + c = 25,x + y + z = 36, ax + by + cz = 30, find the value of x+y+z .
2
2
2
2
2
2
Solution: Cauchy’s Inequality implies
25 × 36 = (a + b + c )(x + y + z ) ≥ (ax + by + cz) = 30× 36 = (a + b + c )(x + y + z ) ≥ (ax + by + cz) = 30 . The equal sign is
25 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2
obtained since 25 × 36 = 30 . Thus there exist λ, μ (not both zero) such that
2
λa = μx, λb = μy, λc = μz . Therefore
2 2
2 2
2 2
λ (a + b + c )= μ (x + y + z ) ⇒ 25λ = 36μ ⇒ 5λ = ±6μλ (a + b + c )= μ (x + y + z ) ⇒ 25λ = 36μ ⇒ 5λ = ±6μ . However,
2 2
2 2
2 2
2 2
2 2
2 2
2 2

μ 5 a+b+c μ 5
ax + by + cz = 30, thus 5λ =6μ ⇒ λ = 6 ⇒ x+y+z = λ = .
6



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110

ELEMENTARY ALGEBRA EXERCISE BOOK I inequAlities




4
s
t
2
2
2
3.70 Let r,s,t satisfy 1 ≤ r ≤ s ≤ t ≤ 4 , find the minimum value of (r − 1) +( − 1) +( − 1) +( − 1) 2
t s r
s
4
t
(r − 1) +( − 1) +( − 1) +( − 1) .
2
2
2
2
t s r
t
s
4
t
s
4
s
2
2
2
2
2
2
Solution: (r−1) +( −1) +( −1) +( −1) ≥ [ (r−1)+( −1)+( −1)+( −1) ] ⇒ 4[(r−1) +( −
s
t
r
t
4
s
(r−1)+( −1)+( −1)+( −1)
2
t 2
t
s
s
s 2
4
r2
2
(r−1) 2 t +( −1) +( −1) +( −1) ≥ [ t t 2 s 2 4 r ] ⇒ 4[(r s −1) +( − t 2
s
2
t
2
4
2
2
4
t
r
t
s
1) +( −1) +( −1) ] ≥ [(r −1)+( −1)+( −1)+( −1)] = [(r + + + )−4]
1) +( −1) +( −1) ] ≥ [(r −1)+( −1)+( −1)+( −1)] = [(r + + + )−4] .
2
2
s 2
r
4
t
s 4
r 2
s
s
t
4
s
2
t
r
t t
s r t s r t s r
√
t
s
s
t
4
Cauchy’s Inequality implies that r + + + 4 ≥ 4 4 r · · · 4 =4 4, thus
t s r t s r
√ √
2 2
2 2
2 2
2 2
4 4
2 2
2 2
√√
s s
2 2
4 4
4[( 4[(r − 1)4[(r − 1) s s t t +( − 1) +( − 1) +( − 1) ] ≥ [4 4 − 4] ⇒ (r − 1) +( − 1) ++( − 22 t t s s +( − 1) +( − 1) ] ≥ [4 4 − 4] ⇒ (r − 1) +( − 1) +

22
22
22
t t
ss
22
ss
44
4[(r − 1) +( − 1) +( − 1) +( − 1) ] ≥ [4 4 − 4] ⇒ (r − 1) +( − 1) +r − 1) +( − 1) +( − 1) +( − 1) ] ≥ [4 4 − 4] ⇒ (r − 1) +( − 1) + 22
2
2
4 4
√ √1)
r r
t t
t t
4 4
t t
√√
2 2
2 2
2 2
( ( − 1) +( − 1) ≥ 4( 2 − 1)− 1) +( − 1) ≥ 4( 2 − 1) ss 22 rr t t
22
t t
s s ( − 1) +( − 1) ≥ 4( 2 − 1)− 1) +( − 1) ≥ 4( 2 − 1) . The equal sign is obtained if and only if
22
44
(
ss r r rr
√ √
r = 2,s =2,t =2 2. Hence, the minimum value of
√
(r − 1) +( − 1) +( − 1) +( − 1) is 4( 2 − 1) .
2
s
4
t
2
2
2
2
t s r
3.71 Real numbers a 1 ,a 2 satisfy a + a ≤ 1, show that for any real numbers b 1 ,b 2,
2
2
2
1
(a 1 b 1 + a 2 b 2 − 1) ≥ (a + a − 1)(b + b − 1) always holds.
2
2
2
2
2
2
1
2
1
Proof: If b + b − 1 > 0, since a + a ≤ 1, we have (a + a − 1)(b + b − 1) ≤ 0, then
2
2
2
2
2
2
2
2
2
1
2
1
1
1
2
2
obviously (a 1 b 1 + a 2 b 2 − 1) ≥ (a + a − 1)(b + b − 1) . If b + b − 1 ≤ 0, Mean Inequality
2
2
2
2
2
2
2
1
2
2
1
1
2
2
2
2
2
2
2
a +b 2 a +b 2 1 2 2 2 2 (1−a −a )+(1−b −b )
2
2
1 2
2
1 2

implies that a 1 b 1 ≤ 1 1 ,a 2 b 2 ≤ 2 2 . Thus a 1 b 1 + a 2 b 2 ≤ (a + a + b + b ) ≤ 1 ⇒ 1 − a 1 b 1 − a 2 b 2 ≥ (1−a −a )+(1−b −b ) 2 ⇒
2
2
2
1
2
1
1
2
2
2 2 a 1 b 1 + a 2 b 2 ≤ 1 2 (a + a + b + b ) ≤ 1 ⇒ 1 − a 1 b 1 − a 2 b 2 ≥ (1−a −a )+(1−b −b ) 2 ⇒
2
2 2
1 2
2
1 2
2
2
2
2
2 2
1
2 1
2 2
1
2
a 1 b 1 + a 2 b 2 ≤ 2 2 (a + a + b + b ) ≤ 1 ⇒ 1 1− a 1 b 1 − a 2 b 2 ≥ 1 2 2 2 2 2 2 ⇒ 2
2 2
(1−a −a )+(1−b −b ) 2
2
] ≥ [(1 − a − a ) + (1 − b − b ) ] ≥ (a +
2 2
2
1
2 2
2
1 2 1
2
2
1 2
2
2
2
2
2
2
(1−a −a )+(1−b −b ) )
2
2
2 2
2
2
2 2
2
2
1
1
1 1 2 2 2 2 2 2 2 2 (1−a −a )+(1−b −b 2 2 2 2 (1 − a 1 b 1 − a 2 b 2 ) ≥ [ (1−a −a )+(1−b −b ) 2 1 2 [(1 − a − a ) + (1 − b − b ) ] ≥ (a +
1
2
⇒
2
1
2
2
1
1 2
1
2
1 2
2 2
1
a 1 b 1 + a 2 b 2 ≤ (a + a + b + b ) ≤ 1 ⇒ 1 − a 1 b 1 − a 2 b 2 ≥
a 1 b 1 + a 2 b 2 ≤ (a + a + b + b ) ≤ 1 ⇒ 1 − a 1 b 1 − a 2 b 2 ≥ ⇒ (1 − a 1 b 1 − a 2 b 2 ) ≥ [ (1−a −a )+(1−b −b ) 2 ] ≥ 1 1 2 2 2 2 2 1 2 2 2 2 1
2
2
2
2
2
1
1
2
2
a − 1)(b + b − 1)
2 2 1 1 2 2 2 2 2 2 1 1 2 2 2 2 2 2 2 2 (1 − a 1 b 1 − a 2 b 2 2 ) ≥ [ 2 2 ] ≥ [(1 − a − a ) + (1 − b − b ) ] ≥ (a +
1
2
1
2
2
2
1
2
2
2
a − 1)(b + b − 1)
1
2
2
2
2
2
1
1
(1−a −a )+(1−b −b ) 2
2
2
2 2
2 2
2
2 2
2 2
(1 − a 1 b 1 − a 2 b 2 ) ≥ [ [
] ≥ [(1 − a − a ) + (1 − b − b ) ] ≥ (a +
2
1
2
2
2
1
1
1
2
2
1
2
2
(1 − a 1 b 1 − a 2 b 2 ) ≥ (1−a −a )+(1−b −b ) 2 2 ] ≥ [(1 − a − a ) + (1 − b − b ) ] ≥ (a + a − 1)(b + b − 1) .

2
1
1
2
1
1
2
2
1
1
2
2
2
2
2
2
1
2
2
2
2
2
2
a − 1)(b + b − 1)
a − 1)(b + b − 1)
1
2
2
2
2
1
1 a log 2
1 x(x−a+1)
3.72 Let A = {x|1+ 1 − 1 < 0}; B = {x|( ) 3 < ( ) ,a ∈ R} ,
log x log x 3 2
3
5
find the range of a such that A ⊆ B.
1
3
−1
1
Solution: 1+ log 3 x − log 5 x < 0 ⇒ 1 + log 3 − 2 log 5 < 0 ⇒ log x 25 < log x . Thus x> 1,
x
x
x
1
1 x(x−a+1)
1 a log 2
then > 3 , then 1 <x< 25 . Hence, A = {x|1 < x< 25 }. ( ) 3 < ( ) log 2 −a < 2 −x(x−a+1) ⇒ 2 −a < 2 −x(x−a+1) ⇒−a<
3
x 25 3 3 1 a log 2 1 x(x−a+1) ⇒ 3 −a −x(x−a+1) −a −x(x−a+1)
3
2
log 2
( ) 3 < ( ) ⇒ 3 3 < 2 ⇒ 2 < 2 ⇒−a<
−x(x − a + 1) ⇒ (x − a)(x + 1) < 0
2
3

1 a log 2
1 x(x−a+1)

( ) 3 < ( ) ⇒ 3 log 2 −a < 2 −x(x−a+1) ⇒ 2 −a < 2 −x(x−a+1) ⇒−a< −x(x − a + 1) ⇒ (x − a)(x + 1) < 0 ( ).
3
3 2
−x(x − a + 1) ⇒ (x − a)(x + 1) < 0
When a = −1, ( ) has no solution.
When a> −1, ( ) has the solution −1 < x <a .
When a< −1, ( ) has the solution a<x< −1.
⎧
⎨ φ (a = −1)
25
Hence, B = {x|− 1 <x <a} (a> −1) from which we know that when a ≥ 3 , A ⊆ B .
{x|a< x< −1} (a< −1)
⎩
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111

ELEMENTARY ALGEBRA EXERCISE BOOK I inequAlities




3.73 x,y,z are positive numbers, show (x+1) 3 + (y+1) 3 + (z+1) 3 ≥ 81 .
y
z
4
x
3
3
Proof: Since x,y,z > 0,Mean Inequality implies that (x+1) + 27 y + 27 ≥ 3 3 (x+1) 3 · 27 y · 27 = 27 (x + 1) ⇒ (x+1) ≥ 27 (x − y)+ 27
y 2 4 y 2 4 2 y 2 4

(x+1) 3 + 27 y + 27 ≥ 3 3 (x+1) 3 27 27 = 27 (x+1) 3 27 (x − y)+ 27 . Similarly, we can obtain (y+1) 3 27 27 (z+1) 3 27 27
y 2 4 y · 2 y · 4 2 (x + 1) ⇒ y ≥ 2 4 z ≥ 2 (y − z)+ 4 , x ≥ 2 (z − x)+ 4
(y+1) 3 27 (y − z)+ 27 (z+1) 3 27 (z − x)+ 27 . Add them up to obtain the aimed inequality.
,
z ≥ 2 4 x ≥ 2 4
√ √
3.74 m, n are positive numbers, show m +1 > n holds if and only if for any
√
x
x> 1, mx + x−1 > n .
360°
√ √
1
x
Proof: m, n > 0,x − 1 > 0, then mx+ x−1 = mx−m+m+ x−1+1 =[m(x−1)+ x−1 ]+m+1 ≥ 2 m+m+1 = ( m+1) 2

x−1
√ √ 1
2
mx+ x = mx−m+m+ x−1+1 =[m(x−1)+ 1 ]+m+1 ≥ 2 m+m+1 = ( m+1) . If and only if m(x − 1) = 1 , i.e. x = 1+ √ , mx + x has
x−1 x−1 x−1 x−1 m x−1
√ √ thinking.
x
2
the minimum value ( m + 1) . Hence, mx + x−1 > n for any x> 1 if and only if
√
( m + 1) >n , i.e. √ m +1 > √ n . 360°
2
thinking.











360°
thinking.



360°
thinking.











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112
112
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ELEMENTARY ALGEBRA EXERCISE BOOK I inequAlities



3.75 Given f(x) = ax + bx , and 1 ≤ f(−1) ≤ 3, 2 ≤ f(1) ≤ 4, find the range of
2
f(−3).


Solution: f(−1) = a − b, f(1) = a + b, f(−3) =9a − 3b . Let f(−3) = mf(−1) + nf(1)

where m, n are parameters ready to be determined. 9a − 3b = m(a − b)+ n(a + b) =(m + n)a − (m − n)b


9a − 3b = m(a − b)+ n(a + b) =(m + n)a − (m − n)b . Comparing the coefficients to obtain m + n =9 ⇒ m =6,n =3. Thus
m − n =3
f(−3) = 6f(−1) + 3f(1) . Since 1 ≤ f(−1) ≤ 3, 2 ≤ f(1) ≤ 4 , we have 12 ≤ 6f(−1) + 3f(1) ≤ 30

12 ≤ 6f(−1) + 3f(1) ≤ 30, then 12 ≤ f(−3) ≤ 30. Therefore the range of f(−3) is [12, 30].

π
3.76 Given 0 <b< 1, 0 <a < , and x = (sin α) log sin α ,y = (cos α) log cos α ,z = (sin α) log cos α
b
b
b
4
x = (sin α) log sin α ,y = (cos α) log cos α ,z = (sin α) log cos α , determine the order of x,y,z .
b
b
b
π
Solution: 0 <b< 1, thus f(x) = log x is a decreasing function. 0 <a < , thus
b
4
0 < sin α< cos α< 10 < sin α< cos α< 1. Therefore, log sin α> log cos α> 0, then

b
b
(sin α) log sin α < (sin α) log cos α ,
b
b
i.e. x< z . And (sin α) log cos α < (cos α) log cos α , i.e. z <y . Hence,
b
b
we obtain the order x <z <y .
3.77 Consider a triangle with side lengths a, b, c , and its area is 1/4, the radius of
√ √ √
1
its circumcircle is 1. If s = a + b + c, t = 1 + 1 + . Compare s and t .
a b c
Solution: Let C be the angle whose opposite side length is c , and the radius of circumcircle
1
R =1, then c =2R sin C = 2 sin C . In addition, ab sin C = . Therefore abc =1. Then
1
2 4
√
√ √
1 1
1
1
1
1
1
1
1 1
1
1
1 1

t = 1 a + + = ( + )+ ( + )+ ( + ) ≥ 11 ab + 11 bc + 11 ca = 11 c+ a+ b √ √


√ √
√ √
= bb
√
c+ a+ a+
c+
1 11
2 c 1
1 11
11
2 b 1
2 a 1
11
11
11
1 11
11
t = =
c
c
b
+ + + = ( + )+ ( + )+ ( + )
b
+
a
2
√ √ t aa √ bb cc = ( + )+ ( + )+ ( + ) ≥ ≥ ab ab + + bc bc + + ca ca = = abc √ √ abc abc = =
cc
2 aa
bb
2 cc
aa
2
2
2 bb
√ √
√
a + √ b + c = s
√ √
a
a + + b + + c = ss . The equal sign can only be obtained if a = b = c = R =1, which is
b
c =
impossible. Hence, s <t .
3 1 1 1
3.78 a, b, c are positive numbers and a + b + c ≤ 3, show ≤ a+1 + b+1 + c+1 < 3.
2
1
1
1
1
1
1
Proof: Since a, b, c > 0, we have a+1 < 1, b+1 < 1, c+1 < 1 , then a+1 + b+1 + c+1 < 3.
Mean Inequality implies
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113

ELEMENTARY ALGEBRA EXERCISE BOOK I inequAlities




1 + 1 + 1 ≥ 3 3 1 , (a+1)+(b+1)+(c+1) ≥ 3 3 (a + 1)(b + 1)(c + 1)
a+1 b+1 c+1 (a+1)(b+1)(c+1) 1 1 1 1 3
Therefore, ( + + )[(a+1)+(b+1)+(c+1)] ≥ 3 3 ·3 (a + 1)(b + 1)(c + 1) =
a+1 b+1 c+1 (a+1)(b+1)(c+1)
( 1 + 1 + 1 )[(a+1)+(b+1)+(c+1)] ≥ 3 3 1 ·3 3 (a + 1)(b + 1)(c + 1) = 9 . Since
a+1 b+1 c+1 (a+1)(b+1)(c+1)
9
1 + 1 + 1 9 9 = 3
0 <a + b + c ≤ 3, then a+1 b+1 c+1 ≥ (a+1)+(b+1)+(c+1) ≥ 3+3 2 .
3.79 Given a, b, c, m, n, p > 0, and a + m = b + n = c + p = R , show an + bp + cm < R 2

an + bp + cm < R .
2
Proof: Construct an equilateral triangle ABC with side length R . Choose points

D,E,F on sides AB, BC, CA respectively such that

AD = a, DB = m, BE = c, EC = p, CF = b, FA = n . In this way, three side lengths are

a + m, c + p, b + n , and a + m = c + p = b + n = R . Connect D with E , connect E

with F , and connect F with D . Let S ADF = S 1 ,S BDE = S 2 ,S CEF = S 3 ,S ABC = S

√
1
1
1
0
0
0
S ADF = S 1 ,S BDE = S 2 ,S CEF = S 3 ,S ABC = S . Then S 1 + S 2 + S 3 = an sin 60 + cm sin 60 + bp sin 60 = 3 (an + cm + bp)
2 2 2 4
√ √
1
1
1
1
0
0
2
0
0
2
S 1 + S 2 + S 3 = an sin 60 + cm sin 60 + bp sin 60 = 3 (an + cm + bp). S = R sin 60 = 3 R . S = S 1 + S 2 + S 3 + S DEF >S 1 + S 2 + S 3 , thus
2 2 2 4 2 4
√ √
3 (an + cm + bp) < 3 R , that is, an + bp + cm < R .
2
2
4 4
2
3.80 Let x 1 ,x 2 , ··· ,x n are positive numbers, show x 2 1 + x 2 2 + ··· + x 2 n−1 + x n ≥ x 1 + x 2 + ··· + x n
x 2 x 3 x n x 1
x 2 x 2 x 2 2
1 + 2 + ··· + n−1 + x n ≥ x 1 + x 2 + ··· + x n .
x 2 x 3 x n x 1
2 2 2 x 2 1
Proof 1: Since x 1 ,x 2 , ··· ,x n > 0 , we can do the following: (x 1 − x 2 ) ≥ 0 ⇒ x + x ≥ 2x 1 x 2 ⇒ x 2 + x 2 ≥ 2x 1

2
1
2 2 2 x 2 1 x 2 2 x 2 n−1 x n 2
(x 1 − x 2 ) ≥ 0 ⇒ x + x ≥ 2x 1 x 2 ⇒ x 2 + x 2 ≥ 2x 1. Similarly, we can obtain x 3 + x 3 ≥ 2x 2 , ··· , x n + x n ≥ 2x n−1 , x 1 + x 1 ≥ 2x n
2
1
2
2
x
2
x
x
. Add them up to obtain ( x 2 1 + x 2 2 + ··· + x 2 n−1 ( + 1 2 x n 2 ) +(x 1 + x 2 + ··· + x n ) ≥ 2(x 1 + x 2 + ··· + x n ) ⇒
+
+ ··· +
+
n−1
2
x n
) +(x 1 + x 2 + ··· + x n ) ≥ 2(x 1 + x 2 + ··· + x n ) ⇒
x 2 2 x 3 2 x n x 2 x 1 x 3 x n x 1
2
x
x
x
+ ··· +
2
n−1
1
≥ x 1 + x 2 + ··· + x n ≥ x 1 + x 2 + ··· + x n.
( x 2 1 + x 2 2 + ··· + x 2 n−1 + x n 2 ) +(x 1 + x 2 + ··· + x n ) ≥ 2(x 1 + + x 2 + ··· + x n ) ⇒ + x x 1 2 2 n + x 2 2 + ··· + x 2 n−1 + x 2 n
x 1
x 2 x 3 x n x 1 x 2 x 3 x n x 2 x 3 x n x 1
x 2 1 + x 2 2 + ··· + x 2 n−1 + x 2 n
Proof 2: Mean Inequality implies that 1 + x 2 ≥ 2 1 · x 2 =2x 1 . Similarly, we have
x 2 x 3 x n x 1 ≥ x 1 + x 2 + ··· + x n x 2 x 2
x 2 x 2
2
x 2 2 + x 3 ≥ 2x 2 , ··· , x 2 n−1 + x n ≥ 2x n−1 , x n + x 1 ≥ 2x n . Add them up to obtain the result.
x 3 x n x 1
2
2
2
2
2
Proof 3: x n + x 2 1 + x 2 1 + x 2 2 + ···+ x 2 n−1 + x n = 2 x n +x 2 1 + x +x 2 2 + ···+ x 2 n−1 2 +x n ≥ 2( x 1x n + x 1 x 2 +
1
2
2
2
2
2
2
2
2
2
2
x
x
+
+
=
+
+
x 1
1
x 2
x n + x n−1 x n + x 2 + ···+ x n−1 x n x n x n x n +x 1 x 1 + x +x 2 x 2 + ···+ x n−1 +x x n n ≥ 2( x 1x n x 1 x 1 x 2 x 2
x 1
1
x 2
1
x n
x n
2 2 2 2 x 2 2 2 2 2 2 x 2 2 ··· + x 1 x 2 ) = 2(x 1 + x 2 + ··· + x n ) x 1 x 2 x n x 1 x 2
x 1
x 2
+ + + + ···+ + = + + ···+ ≥ 2( + + ··· + x n ) = 2(x 1 + x 2 + ··· + x n ) which implies the result.
1
x n x 1 x 1 x 2 n−1 x n x n +x 1 x +x 2 n−1 +x n x 1x n x 1 x 2 x n−1 x n
x 1 x 1 x 2 x 2 x n x n x 1 x 2 x n x 1 x 2 x n
··· + x n−1 x n ) = 2(x 1 + x 2 + ··· + x n )
x n
Proof 4: x 2 1 + x 2 2 + ··· + x 2 n−1 + x n 2 = [x 2−(x 2 −x 1)] 2 + [x 3−(x 3 −x 2)] 2 + ··· + [x n−(x n−x n−1)] 2 +.
x 2 x 3 x n x 1 x 2 x 3 x n
[x 1−(x 1 −x n)] 2
=(x 2 + x 3 + ··· + x n + x 1 ) − 2(x 2 − x 1 + x 3 − x 2 + ··· + x n − x n−1 + x 1 −
x 1 2 2 2 2 2
x n )+ (x 2 −x 1) + (x 3 −x 2) + ··· + (x n−x n−1) + (x 1 −x n) =(x 1 + x 2 + ··· + x n )+ (x 2 −x 1) +
x 2 x 3 x n x 1 x 2
(x 3 −x 2) 2 (x 1 −x n) 2
+ ··· + ≥ x 1 + x 2 + ··· + x n
x 3 x 1
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114

ELEMENTARY ALGEBRA EXERCISE BOOK I inequAlities



√ √ √ √
Proof 5: Since x 1 ,x 2 , ··· ,x n > 0, let a 1 = x 2 ,a 2 = x 3 , ··· ,a n−1 = x n ,a n = √ ,b 2 = x 2 √ ,b 3 =
x 1
√
√
x 2
√
√
x 2 =
x 3 =
a 1 = x 2 ,a 2 = x 3 , ··· ,a n−1 = x n ,a n = x 1 ,b 1 = √ ,b 2 √ ,b 3
x 1 ,b 1 = x 1
x 3 x n−1 x n x 2 x 3
√ √ √ √ √ , ··· ,b n−1 = √ ,b n = √
x n−1
x 3
a 1 = x 2 ,a 2 = x 3 , ··· ,a n−1 = x n ,a n = x 1 ,b 1 = √ ,b 2 = √ ,b 3 = √ , ··· x 4 ,b n−1 = √ ,b n x n = √ x n . Cauchy Inequality implies that
x 2
x 1
x 1
x 2 x 3 x 4 x n x 1
x 3 x n−1 x n
√ , ··· ,b n−1 = √ ,b n = √ √ 2 √ 2 √ 2 √ 2 2 2 x n−1 2 2
x n
x 4 x n x 1 2 2 2 2 2 2 2 √ √ √ √ x 1 x 2 ) +( √ ) ] ≥
(a + a + ··· + a )(b + b + ··· + b ) ≥ (a 1 b 1 + a 2 b 2 + ··· + a n b n ) , then [( x 2 ) +( x 3 ) +···+( x n ) +( x 1 ) ]·[( √ ) +( √ ) +···+( √ 2
2
2
2
2
2
x n−1 2
2
√
√
x n
x 2
√
x 1
√
1 2 n 1 2 n [( x 2 ) +( x 3 ) 2 +···+( x n ) +( x 1 ) ]·[( √ x 1 x 2 ) +( √ ) +···+( √ x n ) +( √ ) ] ≥
x n−1 2
2
x 3 2
2
2
x 1 2
2
x 2
x n
2 x 1) ] ≥
√ [( x 2 ) +( x 3 ) +···+( x n ) +( x 1 ) ]·[( √ ) +( √ ) +···+( √ x n) +( √ x x 1 x 2
√
x 3
√
x 2
√
2
x n−1
√ √ 2 √ √ 2 2 2 x n−1 2 2 √ √ x 1 + √ x 3 x 2 √ x 2 + ··· + √ x n x n−1 + √ x 1 x n x n x 2 x 3 x n 2 2 1 + x 2 2 2 +
√ ] ⇒ (x 2 + x 3 + ··· + x n + x 1 )( x
√
2
√
x n
[ x 2 x 1
x 1

x 2

x
2 √ √
2 √ √
√ √
√ √
2
√
√
√
√
x
x
[( x 2 ) +( x 3 ) +···+( x n ) +( x 1 ) ]·[( √ ) +( √ ) +···+( √ ) +( √ ) ] ≥ [ x 2 x 1 x 2 + √ x 3 x 2 x 3 + ··· + √ x n n−1 x n + √ x 1 x n x 1 ] ⇒ (x 2 + x 3 + ··· + x n + x 1 )( x 1 1 x 2 + x 2 2 x 3 +
x
2 2
2 2
2 2
2 2
2 2
x n−1 2 2
2 2
2 2
x n−1
x 3 ) +···+(
x 1
x n n
x 2 2
[( x 2 ) +( x 3 ) +···+( x n ) +( x 1 ) ]·[( √ 1 x 2 ) +( √ ) +···+( √ √ x n ) +( √ ) ] ≥ [ x 2 √ x 2+ x 3 √ x 3+ ··· + x n √ x n + x 1 √ x 1] ⇒ (x 2 + x 3 + ··· + x n + x 1 )( x 2+ x 3+
2
x 1 ) ] ≥
[( x 2 ) +( x 3 ) +···+( x n ) +( x 1 ) ]·[( √ ) +( √
2
x
) +( √
n
≥ (x 1 + x 2 + ··· + x n−1 + x n )
x 3 3
x 2 2
x n n
√ √ √ √ x x x x x x x 2 ··· + x x 2 2 n−1 + x 2 x x 3 x n x 1 2 x 2 x 3
2 1 1
2
+
x n
√
√
√ √ √ x 1 + √ √ x 2 + ··· + √ x n−1 √ √ √ ] ⇒ (x 2 + x 3 + ··· + x n + x 1 )( x x 2 2 1 + x x 2 2 2 x 2 n−1 2 n x 1 ≥ (x 1 + x 2 + ··· + x n−1 + x n ) 2 2
x n +x
x
2 2

√
x
+ ··· + n−1
x n n−1
n
+
[ x 2 x 1 1
[ [ x 2 x 2 √ √ x x 2 + + x 3 x 3 x 3 x 2 2 x 3 + ··· + + x n x n x x n−1 x n+ + x 1 x 1 x 1 x n n x 1 ] ⇒ (x 2 + x 3 + ··· + x n + x 1 )( )( 1 1 x 2 + + 2 2 x 3 + ··· + x n + x 1 ≥ (x 1 + x 2 + ··· + x n−1 + x n )
+ ···
√ √
√ √
√ √ ] ⇒ (x 2 + x 3 + ··· + x n + x 1
x x 2 2 2 x 2 x x x x x x n x 1
x n n
x 1 1
x 3 3
x 2 2
x 3 3
x
n−1
··· + x x 2 2 n−1 + x x 2 2 n ≥ (x 1 + x 2 + ··· + x n−1 + x n ) 2 2 2
··· + + n−1 n n x 1≥ (x 1 + x 2 + ··· + x n−1 + x n ) ) . Divide both sides by x 1 + x 2 + ··· + x n−1 + x n ≥ 0 to obtain
x n + +
···
≥ (x 1 + x 2 + ··· + x n−1 + x n
x x
x n n
x 1 1
x 2 1 + x 2 2 + ··· + x 2 n−1 + x n 2 .
x 2 x 3 x n x 1 ≥ x 1 + x 2 + ··· + x n−1 + x n
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115
115

ELEMENTARY ALGEBRA EXERCISE BOOK I inequAlities




3.81 If x, y are real numbers, and y ≥ 0,y(y + 1) ≤ (x + 1) , show y(y − 1) ≤ x .
2
2
2
2
2
2
1 2
2
Proof: If 0 ≤ y ≤ 1, obviously y(y − 1) ≤ 0 ≤ x . If y> 1, then y(y + 1) ≤ (x + 1) ⇒ y + y + 1 ≤ (x + 1) + 1 ⇒ (y + ) ≤ (x + 1) + 1 ⇒ 1 <
2
2
2
4
2
1 2
2
4
y(y + 1) ≤ (x + 1) ⇒ y + y + 1 ≤ (x + 1) + 1 ⇒ (y + ) ≤ (x + 1) + 1 ⇒ 1 < 4
2 1 1 4 4 2 4

−
1
2
2
2
1 2
2
2
(x + 1) +
y(y + 1) ≤ (x + 1) ⇒ y + y + 1 ≤ (x + 1) + 1 ⇒ (y + ) ≤ (x + 1) + 1 ⇒ 1 < y ≤ y ≤ (x + 1) + 1 2 4 . The inequality
2
−
4
4 4 2 4
2
y ≤ (x + 1) + 1 − 1 to prove 2 2 2 2 2 2 1 1 1 2 2 2 1 1 1 ⇔ (y − ) ≤ x + 1 1 2 1 1 1 1 1 1
1 2 2
2
1 2
1
22 1
2 2
1
1
1
2
y(y − 1) ≤ x ⇔ y − y + ≤ x + ⇔ (y − ) ≤ x + 2
2
1 2
4 2 y(y − 1) ≤ x ⇔ y − y + 4 + 1 4 x + 2 4 1 4 4⇔ (y − ) ≤ x + 4 4 ⇔ y ≤ x + x + + 2 ⇔ ⇔
x
4+
y(y − 1) ≤ x ⇔ y − y
4 2 +
4 ≤ x +
⇔ y ≤ y ≤ x + 2
⇔
+ + ⇔
y(y − 1) ≤ x ⇔ y − y + ≤
2 2
2 ⇔
≤ x + ⇔ (y − ) ≤ x + ⇔ y ≤
2
4
4
4
2
4
2
1
2
1
1
2

1− y +
2 ) ≤ x +
y(y − 1) ≤ x 2 1 ⇔ y 1 4 1 ≤ x + 1 2 2 4 x + 2 + 1 2
⇔ (y −
1
(x + 1) + − x + + 1 1 ⇔ 4 (x + 1) + 1 ≤ 2 ⇔ y ≤ 4 2 1 ⇔
1
1
≤ 1
1
2 1
2 2
2 1
2 1 1
2 1
2 4
≤ 2 14
2 1 1
x + +1 ⇔ (x + 1) + 2 1 2
2
2
1
1
1
x + 1
1
4 +1
x
4 +
(x + 1)
x2
(x + 1)
2
4 2 +
2
x
4 +
(x + 1) + ≤
(x + 1) + − 4 ≤ x + + 1 2 ⇔ (x + 1) + 2 4 ≤ x + +1 ⇔ (x + 1) + 2 4 1 ≤
(x + 1) + − ≤
4 + +1 ⇔ (x + 1) + ≤
2 ⇔
2 ≤
4 4 2 − 2 2 4 + 4 4 + ⇔ 4 4 + 4 ⇔ (x + 1) + 4 ≤
4
4 ≤
2
4
1
1
1
1
1
2 2
1 2
1 2
1 2
2 2
x + +1
x + +

(x + 1)
(x + 1) + 1 x + +1 ⇔ x +2x +1 ≤ x +2 + x + +1 ⇔ x ≤ ⇔ (x + 1) + 1 ≤

1
2
1
⇔
2 ≤
x +
2 ≤
2
1
1
1
1
2 1 2
2 2 1 x + +2 − 1 1 4 2 2 4 2 1 1 1 4 2 2 4 2 2 1 4 +1 ⇔ x ≤ x + 4 + 1 1 4
2 1
2 4
4 1 4
2
1
x + 1
1
2
2
x
2
4 +
4 +1 ⇔ x +2x +1 ≤ x +2
x 2 4 +2 4 x + +1 ⇔ x +2x +1 ≤ x +2 x + +1 ⇔ x ≤ x
x
x2
2
4
x + +2 x + +1 ⇔ x +2x +1 ≤ x +2
x + +2 4 + 4 4 4 4 4 + 4 +1 ⇔ x ≤ 4 + 4 4
4
4
4
4
1
1
2
1
2
1
2
1
2
2
2
x + +2 x + +1 ⇔ x +2x +1 ≤ x +2 x + +1 ⇔ x ≤ x + 1 which is
4 4 4 4 4 4
obviously valid.
3.82 If real numbers x,y,z satisfy x + y + z =2, show x + y + z ≤ xyz +2.
2
2
2
Proof: If one (or more) of x,y,z is not positive, without loss of generality let z ≤ 0. Since
1 2 2 1 2 2 2
2
2
2
2
2
x+y ≤ 2(x + y ) ≤ 2(x + y + z ) =2, xy ≤ (x +y ) ≤ (x +y +z )= 1, then
2 2
2+ xyz − (x + y + z) = [2 − (x + y) − z(xy − 1)] ≥ 0, that is, x + y + z ≤ xyz +2.
If x,y,z are all positive, let 0 <x ≤ y ≤ z .
When z ≤ 1, 2+ xyz − (x + y + z)=1 − x − y + xy +1 − xy − z + xyz = (1 − x) − y(1 − x)+

2+ xyz − (x + y + z)=1 − x − y + xy +1 − xy − z + xyz = (1 − x) − y(1 − x)+
(1 − xy) − z(1 − xy) = (1 − x)(1 − y) + (1 − xy)(1 − z) ≥ 0

2+ xyz − (x + y + z)=1 − x − y + xy +1 − xy − z + xyz = (1 − x) − y(1 − x)+ (1 − xy) − z(1 − xy) = (1 − x)(1 − y) + (1 − xy)(1 − z) ≥ 0 , that is,x + y + z ≤ xyz +2

(1 − xy) − z(1 − xy) = (1 − x)(1 − y) + (1 − xy)(1 − z) ≥ 0
x + y + z ≤ xyz +2.
√
2
2

When z> 1, x + y + z ≤ 2[z +(x + y) ]= 2(2 + 2xy)= 2 1+ xy ≤ 2+ xy < 2+ xyz
√
2
2
x + y + z ≤ 2[z +(x + y) ]= 2(2 + 2xy)= 2 1+ xy ≤ 2+ xy < 2+ xyz .
As a conclusion, x + y + z ≤ xyz +2 holds.
3.83 Given the function f(x) = ax + bx + c (a> 0), and the two roots of the
2
1
equation f(x) − x =0 satisfy 0 <x 1 <x 2 < . (1) When x ∈ (0,x 1 ), show x< f(x) <x 1 ;
a
(2) Assume the curve of the function f(x) is symmetric about the straight line x = x 0, show
x 0 < x 1 .
2
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116

ELEMENTARY ALGEBRA EXERCISE BOOK I inequAlities




Proof: (1) Let G(x) = f(x) − x . Since x 1 ,x 2 are the two roots of the equation f(x) − x =0,

then G(x) = a(x − x 1 )(x − x 2 ). When x ∈ (0,x 1 ), since x 1 <x 2 ,a > 0, then G(x) = a(x − x 1 )(x − x 2 ) > 0 ⇒ f(x) − x> 0 ⇒ f(x) >x

G(x) = a(x − x 1 )(x − x 2 ) > 0 ⇒ f(x) − x> 0 ⇒ f(x) >x. x 1 − f(x)= x 1 − [x + G(x)] = x 1 − x − a(x − x 1 )(x − x 2 )=(x 1 − x)[1 + a(x − x 2 )]

1
x 1 − f(x)= x 1 − [x + G(x)] = x 1 − x − a(x − x 1 )(x − x 2 )=(x 1 − x)[1 + a(x − x 2 )] . Since 0 <x 1 <x 2 < , we have x 1 − x> 0, 1+ a(x − x 2 ) = 1+ ax − ax 2 > 1 − ax 2 > 0
a
x 1 − x> 0, 1+ a(x − x 2 ) = 1+ ax − ax 2 > 1 − ax 2 > 0, thus x 1 >f(x).

b
(2) x 0 = − . Since x 1 ,x 2 are the roots of the equation f(x) − x =0 , that is,
2a
x 1 ,x 2 are the roots of the equation ax +(b − 1)x + c =0 . Vieta’s formula implies
2
b−1 b a(x 1 +x 2)−1 ax 1 +ax 2−1
x 1 + x 2 = − , thus b =1 − a(x 1 + x 2 ), then x 0 = − = = . Since
a 2a 2a 2a
ax 2 < 1, that is, ax 2 − 1 < 0, then x 0 = ax 1 +ax 2 −1 < ax 1 = x 1 .
2a 2a 2
2
2
2
a +b 2 b +c 2 c +a 2 a 3 b 3 c 3
3.84 Let a, b, c are positive numbers, show a + b + c ≤ + + ≤ + +
2c 2a 2b bc ca ab
2
2
2
a +b 2 b +c 2 c +a 2 a 3 b 3 c 3
a + b + c ≤ + + ≤ + + .
2c 2a 2b bc ca ab
2 1
1
2
2
Proof: Without loss of generality, assume a ≥ b ≥ c> 0, then a ≥ b ≥ c , ≥ 1 ≥ , then
c b a
1
1
1
1
1
1
1
1
1
1
1
1
a · + b · + c · ≤ a · + b · + c · ,a · + b · + c · ≤ a · + b · + c · . Add
2
2
2
2
2
2
2
2
2
2
2
2
a b c b c a a b c c a b
2
2
2
a +b 2 b +c 2 c +a 2 3 3 3 1 1 1
these two inequalities up to obtain a + b + c ≤ + + . a ≥ b ≥ c , ≥ ≥ ,
2c 2a 2b bc ca ab
3 1 3 1 3 1 3 1 3 1 3 1 3 1 3 1 3 1 3 1 3 1 3 1
then a · +b · +c · ≥ a · +b · +c ,a · +b · +c · ≥ a · +b · +c . Add
bc ca ab ab bc ca bc ca ab ca ab bc
2
2
2
these two inequalities up to obtain a 3 + b 3 + c 3 ≥ a +b 2 + b +c 2 + c +a 2 . Hence,
bc ca ab 2c 2a 2b
2
2
2
a +b 2 b +c 2 c +a 2 a 3 b 3 c 3 .
a + b + c ≤ + + ≤ + +
2c 2a 2b bc ca ab
3.85 Solve the inequality log (x +3x +5x +3x + 1) > 1 + log (x + 1).
12
6
4
10
8
2
2
6
4
6
12
12
10
10
8
8
Solution: The inequality is equivalent to log (x +3x +5x +3x + 1) > log (2x + 2) ⇔ x +3x +5x +3x +1 >
2
12
10
8
2 12
6
6
10
8
4
4 log (x +3x +5x +3x + 1) > log (2x + 2) ⇔ x
8
12
10
4
2x +2 ⇔ 2x +1 <x +3x +5x +3x 6 +3x +5x +3x +1 >
2
2
6
8
4
10
12
10
12
6
8
6
4
12
10
4
8
log (x +3x +5x +3x + 1) > log (2x + 2) ⇔ x +3x +5x +3x +1 > 2x +2 ⇔ 2x +1 <x +3x +5x +3x . Obviously
2
2
4
10
8
12
4
2x +2 ⇔ 2x +1 <x +3x +5x +3x 6
x =1 does not satisfy the inequality, thus we can divide both sides by x to obtain
6
2 2 1 1 6 6 4 4 2 2 6 6 4 4 2 2 2 2 2 2 3 3 2 2
1 1
2 2
2 2
4 4
2 2
2 2
6 <x +3x 4 4
6 6
2 2
+5x +3 = x +3x +3x +1+2x +2 = (x +1) +2(x +1) ⇔
6 6
3 3
x 2 + 2 2 x + 2 6 <x +3x +5x +3 = x +3x +3x +1+2x +2 = (x +1) +2(x +1) ⇔
6 <x +3x +5x +3 = x +3x +3x +1+2x +2 = (x +1) +2(x +1)
1 + +
x
x
6 <x +3x +5x +3 = x +3x +3x +1+2x +2 = (x +1) +2(x +1) ⇔ ⇔
2 2
1
2
3
2
3
1 x
( ) + 2( ) < (x + 1) + 2(x + 1)
3 x x
2
3
2
x
1
2( ) + 2( ) < (x + 1) + 2(x + 1)
2 2
3 3
2 1 1
2 1 1
3 3
3
2 2
x
2
x 2( ) < (x + 1) + 2(x +
x x ( ) + 2( ) < (x + 1) + 2(x + 1) 1). Let g(t)= t +2t , then the inequality becomes
(
2 2 ) +
x x x x
2 2

1
g( ) <g(x + 1). Since g(t)= t +2tg(t)= t +2t is an increasing function, then the inequality is
2
3 3
x 2
√
equivalent to 1 1 2 2 2 22 2 2 2 2 5−1 (the other
2 <x +1 ⇔ (x ) + x − 1 > 0<x +1 ⇔ (x ) + x − 1 > 0 whose solution is x >
x x 2 2
√
2 5+1
part x < − is deleted). Hence, the original inequality has the solution set
2

√ √
5−1 5−1 , +∞).
(−∞, − ) ∪ (
2 2
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117

ELEMENTARY ALGEBRA EXERCISE BOOK I inequAlities




3.86 Let a, b, c are integers and at least one of them is nonzero, and their
√ √
6
absolute values are not greater than 10 , show |a + b 2+ c 3| > 10 −21 .
Proof: When b =0,c =0, the conclusion is obviously valid. When one of b, c is nonzero, we consider
√ √ √ √ √ √ √ √
the following four numbers: t 1 = a + b 2+ c 3,t 2 = a + b 2 − c 3,t 3 = a − b 2+ c 3,t 4 = a − b 2 − c 3

√ √ √ √ √ √ √ √ √ √ √
√ 2 2 √ 2 2 2 √ 2 2 2
2
2
2
2
2
2
t 1 = a + b 2+ c 3,t 2 = a + b 2 − c 3,t 3 = a − b 2+ c 3,t 4 = a − b 2 − c 3. They are all irrational numbers, and t = t 1 t 2 t 3 t 4 = [(a + b 2) − 3c ][(a − b 2) − 3c ]= (a +2 2ab +2b − 3c )(a −
2
2√
√
√
√
√
√ t = t 1 t 2 t 3 t 4 = [(a + b 2) − 3c ][(a − b 2) − 3c ]= (a +2 2ab +2b − 3c )(a −
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
√
√ √ √ 2 2ab +2b − 3c ) = [(a +2b − 3c ) +2 2ab][(a +2b − 3c ) − 2 2ab] = [(a +
√
2 = [(a + b 2) − 3c ][(a − b 2) − 3c ]= (a +2 2ab +2b − 3c )(a −
√ t = t 1 t 2 t 3 t 4
2
2
2
2
2
2
2
2 √
2 √
√ 2 2 √ 2 2 2 √ 2 2 2 √ 2 2ab +2b − 3c ) = [(a +2b − 3c ) +2 2ab][(a +2b − 3c ) − 2 2ab] = [(a +

t = t 1 t 2 t 3 t 4 = [(a + b 2) − 3c ][(a − b 2) − 3c ]= (a +2 2ab +2b − 3c )(a −
2
2
2
2
2
2
2
2
2
√t = t 1 t 2 t 3 t 4 = [(a + b 2) − 3c ][(a − b 2) − 3c ]= (a +2 2ab +2b − 3c )(a − 2 2ab 2 2 2 2 2 2 2 2 2 2 2 2
2 +2b − 3c ) = [(a +2b − 3c ) +2 2ab][(a +2b − 3c ) − 2 2ab] = [(a +
√
√
2b − 3c ) − 8a b ] ∈ Z
2
2
2
2 2
2
2
2
2
2
2 2
2
2
2 2ab +2b − 3c ) = [(a +2b − 3c ) +2 2ab][(a +2b − 3c ) − 2 2ab] = [(a +
√
√
√
2 2ab +2b − 3c ) = [(a +2b − 3c ) +2 2ab][(a +2b − 3c ) − 2 2ab] = [(a + 2b − 3c ) − 8a b ] ∈ Z . Thus |t|≥ 1,
2b − 3c ) − 8a b ] ∈ Z
2 2
2
2
2 2
2
2
2
2
2
2
2
2
2 2
2 2
2
2b − 3c ) − 8a b ] ∈ Z √ √
2 2
2
2 2
1
6
2b − 3c ) − 8a b ] ∈ Z which implies that |t 1 |≥ |t 2 |·|t 3|·|t 4| . In addition, since 1+ 2+ 3 < 10 and |a|, |b|, |c|≤ 10 ,
√ √
7
6
we have |t i |≤ (1 + 2+ 3) · 10 < 10 , thus |t 1 | > 1 7 = 10 −21 .
7
7
10 ·10 ·10
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advising and
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helping
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helping foremen foremen advising and
helping foremen
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118
118

ELEMENTARY ALGEBRA EXERCISE BOOK I inequAlities




80
1
3.87 For k ∈ N , show 16 < √ < 17.
k
k=1
√ √ √ √ √ √ √ √
Proof: From k − 1 < k< k +1, we have k + k − 1 < 2 k< k +1+ k .
k ∈ N , then
√ √ √ √ √ √
1 √ √ 1 √ √ 1√ √
√ √ < 12 k< √ 1 √ ⇒ 2( k +1 − k) < √ < 2( k − k − 1) ⇒
1
1
k+1+ k √ < 12 k< √ k+ k−1 ⇒ 2( k +1 − k) < √ < 2( k k − k − 1) ⇒
√
√
k+1+ k n k+ k−1 k
n
√ √ √ √ 1 1 √ √ √ √ , where 1 ≤ m ≤ n , and m, n ∈ N . Choose

2( n +1
m)
√ < 2( n −m − 1)
2( n +1 − − m) < < √ < 2( n − m − 1)
k k
k=m 80
k=m
1
n = 80,m =1, then 16 < √ . Choose n = 80,m =2, then
k
80 80 k=1 80
1 1 √ √ √ √ 1

1+
1+ √ < 2( 80 − 1) + 1 < 2 81 − 1 = 17 17. Hence, 16 < √ < 17.
√ < 2( 80 − 1) + 1 < 2 81 − 1 =

k k k
k=2
k=2 k=1
) .
3.88 For n ∈ N ,n > 1, show n! < ( n+1 n
2
Proof 1: (applying mean inequality)
n!=1 · 2 · ··· · k · ··· · (n − 1) · n (i),
n!= n · (n − 1) · ··· · (n − k + 1) · ··· · 2 · 1 (ii).
2
) =(
) =
(i)× (ii): (n!) = (1 · n)[2(n − 1)] · ··· · [k(n − k + 1)] · ··· · [(n − 1)2](n · 1). Since 1 · n ≤ ( 1+n 2 2+n−1 2 1+n 2 k+n−k+1 2
) , 2(n − 1) ≤ (
) , ··· ,k(n − k + 1) ≤ (
2 2
2
2
2 2
1+n 2 ,k(n − k + 1) ≤ (
) =
)
1
n−1+2 2 (
1+n 2 · n ≤ ( 1+n ) , 2(n − 1) ≤ 2+n−1 2 1+n ) , ··· k+n−k+1 2
1+n 2 =(
( ) , ··· , (n − 1)2 ≤ ( ) =( ) ,n · 1 ≤ ( ) 2
2
2
2
2
2 2
2 2
2 2
) =(
) , ··· ,k(n − k + 1) ≤ (
1 · n ≤ ( 1+n 2 2+n−1 2 1+n 2 k+n−k+1 2 ( 1+n ) , ··· , (n − 1)2 ≤ ( n−1+2 2 1+n ) ,n · 1 ≤ ( 1+n )
) , 2(n − 1) ≤ (
) =(
) =

2−1 2
2n 2
2
2n 2
)
n−1+2 2(
1+n 2 ,k(n − k + 1) ≤ (
) =
1+n 2 · n ≤ ( 1+ ) , 2(n − 1) ≤ 2+n 1+n 2=( 1+ ) , ··· k+n−k+1 2 2 2 2 2
1
( ) , ··· , (n − 1)2 ≤ ( ) =( 2 ) ,n · 1 ≤ ( ) 2
2
2
2
2
2n 2
2n 2
) ] ⇔ (n!) ≤ (
)
) . There are n terms. Thus (n!) ≤ [(
)
) =(
( 1+ ) , ··· , (n − 1)2 ≤ ( n−1+2 2 1+ ) ,n · 1 ≤ ( 1+n 2 2 1+n 2 n 2 n+1 2n ⇔ n! ≤ ( n+1 n
2 2 2 2 2 2 2
2
) .
) . Since n> 1, then n! < (
) ] ⇔ (n!) ≤ (
(n!) ≤ [( 1+n 2 n 2 n+1 2n ⇔ n! ≤ ( n+1 n n+1 n
)
2 2 2 2
Proof 2: (applying mathematical induction)
) = . The inequality holds.
When n =2, LHS=2! =2, RHS=( 2+1 2 9
2 4
) . Then
Assume the inequality holds for n = k , that is, k! < ( k+1 k
2
k+1
k+1
) (k + 1)) (k + 1).
k!(k + 1) < (k!(k + 1) < ( k+1 kk k+1 kk
) (k + 1) ⇔ (k + 1)! < () (k + 1) ⇔ (k + 1)! < (
2 2 2 2
k+1 k k+1
k+1 k+1
(k+1)+1 k+1
( k+1 k (k+1)+1 k+1 ⇒ 2( k+1 k ) < [ (k+1)+1 k+1 ⇒ 2( k+1 k k+1 (k+1)+1 k+1 ⇒ k+1 k+1 (k+1)+1 k+1
< [
]
) (k+1) < [
) (
]
]
)
(k+1)+1 k+1
2 2 ( 2 ) (k+1) < [ 2 ] ⇒ 2( 2 ) ( ) < 2[ ] ⇒ 2( ) < [ ] ⇒
2
(k+1)+1 k+1 2 ) k+1 2 2 2 2 2 2 2
]
( (k+1)+1 k+1
]
]
⇒ 2(
2 )
) (
]
) (k+1) < [
( k+1 k (k+1)+1 k+1 ⇒ 2( k+1 k k+1 ) < [ (k+1)+1 k+1 2 < [ k+1 k+1 < [ k+1 ] ⇒ 2 < [ (k+1)+1 k+1 ( 2 ) k+1 . Binomial theorem implies that
2 2 2 2 2 2 2 2 k+1
]
2 < [ (k+1)+1 k+1 ( 2 ) k+1 1 k+1 1 k+1 kk (k+1)+1 k+1k+1
(k+1)+1
k+1
) (k + 1)
2 k+1 (1 + ) = 1 +(k + 1) + ··· > 2. Thus (( ) (k + 1) < [ < [ ]] holds.
k+1 k+1 2 2 2 2
) holds according to mathematical induction
Hence, the original inequality n! < ( n+1 n
2
above.
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119

ELEMENTARY ALGEBRA EXERCISE BOOK I inequAlities




1
2
3.89 Let a 1 ,a 2 , ··· ,a n be a permutation of 1, 2, ··· ,n , show + + ··· + n−1 ≤ a 1 + a 2 + ··· + a n−1
2
n
3
a 3
a n
a 2
1 + + ··· + n−1 a 1 + a 2 + ··· + a n−1 .
2
2 3 n ≤ a 2 a 3 a n
Proof: Since a 1 ,a 2 , ··· ,a n is a permutation of 1, 2, ··· ,n , we have
(a 1 + 1)(a 2 + 1) · ··· · (a n−1 + 1) ≥ (1 + 1)(2 + 1) · ··· · (n − 1 + 1) = a 1 a 2 ··· a n a n . Thus
(a 1 + 1)(a 2 + 1) · ··· · (a n−1 + 1) ≥ (1 + 1)(2 + 1) · ··· · (n − 1 + 1) = a 1 a 2 ···
2
1a 1 +1
a 1 +1
1 a
1 + + ··· + n−1 a n−1 a 1 + + +···+ + a n−1n−1 1 a 1 1 1 a 2 1 a 1 a n−1n−1 111 1 1 + a 2 +1 1 1 1 a n−1 +1 + a 2 1 +1 1 1 + a 1 +1+1 a 2 +1 1 a n−1 +1 a n−1 +1
+ ···+
+ + +···++ +···++ + +···+
a n−1
+ + +···+
=
a n−1
a n−1
a 1
1a 2
a 2
a 1
1 a 2
=+···+ 1
+ ≥1
+···++ + +···+= 1
+···+ a 1 +1
+ 1
++ +···+ a n−1
≥ a n−1
=+ + +···++ a 1
a 2
2 3 a 2 + +···+≤ a n a 2 1 a 3 2a 3 n a n a a 12 1 a2 a n a nn + 2a+···+ a 3 a n a n + + +···+ a n 3 + + +···+ a 3 = + a n + ≥ a 2 +1 +···+ ≥
a
n
a
a 1 1a 1
a 23
a 3
a 1 a 1
a 22
a 2
a 2
a n a 2
a 1
a 2 a n
a n a n a 2 a 3 a n 1 2 a n a 1 a 2 a n a 1 a 2 a 3 a n

n (a 1 +1)(a 2 +1)·····(a n−1 +1)+1) a n−1 +1
a 2 +1
1
1
1
a 1 +1
1
n
+ +···+ + + +···+ n + + +···+ n= + n +···+ a 1 a 2 ···a n n n (a 1 +1)(a ≥ n . In
≥ n +
≥ n 2 +1)·····(a n−1 +1)
a 1 a 2 a n−1 1 1 a n−1 n (a 1 +1)(a 2 +1)·····(a n−1 +1)(a 1 +1)(a 2 +1)·····(a n−1 ≥ n
≥
a
a 2 a 3 a n 1 2 a n a 1 a 2 a 1 a 2 ···a na n a 1 a 2 a 1 a 2 ···a n 3 a n
a 1 a 2 ···a n
1
2
1
1
1
n n (a 1 +1)(a 2 +1)·····(a n−1 +1) ≥ n addition, n =( + + ··· + ) + ( + + ··· + n−1 ). Hence,
a 1 a 2 ···a n 1 2 n 2 3 n
a
11 + + ··· ++ + ··· + n−1n−1 a 1 a 1 ++ a 2 a 2 + ··· ++ ··· + a n−1n−1 .
22
2 2 3 3 n n ≤≤ a 2 a 2 a 3 a 3 a nn
a
3.90 If real numbers a, b, c satisfy a + b + c =3, show
1 + 1 + 1 ≤ .
1
2
2
2
5a −4a+11 5b −4b+11 5c −4c+11 4
9
Proof: If a, b, c are all less than , then we can show 2 1 ≤ 1 (3 − a) ( ). Actually
5 5a −4a+11 24 2 3 2 2
( ) ⇔ (3−a)(5a −4a+11) ≥ 24 ⇔ 5a −19a +23a−9 ≤ 0 ⇔ (a−1) (5a−9) ≤
2
2
2
3
( ) ⇔ (3−a)(5a −4a+11) ≥ 24 ⇔ 5a −19a +23a−9 ≤ 0 ⇔ (a−1) (5a−9) ≤ 0 ⇔ a< 9 5

3
2
2
2
( ) ⇔
0 ⇔ a< 9(3−a)(5a −4a+11) ≥ 24 ⇔ 5a −19a +23a−9 ≤ 0 ⇔ (a−1) (5a−9) ≤
5
0 ⇔ a< . Similarly, we can obtain 2 1 ≤ 1 (3 − b), 2 1 ≤ 1 (3 − c) . Add these
9
5 5b −4b+11 24 5c −4c+11 24
1
1
1
1
1
1
1
three inequalities up to obtain 1 5a −4a+11 1 5b −4b+11 1 5c −4c+11 1 ≤ 24 (3−a)+ (3−b)+ (3−c) = 24 [9−(a+b+c)] =
+
+

2
2
2
1
24
1
1
24
2
1 2
24
24
5a −4a+11 + 5b 2 1 −4b+11 + 5c −4c+11 ≤ 24 (3−a)+ (3−b)+ (3−c) = 24 [9−(a+b+c)] =
[9 − 3] =
1
1 + 1 + 1 1 (3−a)+ (3−b)+ (3−c) = 1 [9−(a+b+c)] = 1 [9 − 3] = 1 . 4
1
24
2
2
2
5a −4a+11 5b −4b+11 5c −4c+11 ≤ 24 24 24 24 24 4
1 [9 − 3] = 1
24 4
9
9
If at least one of a, b, c is not less than , without loss of generality, assume a ≥ , then
5 5
9
9
5a − 4a + 11 = 5a(a − ) + 11 ≥ 5 · · ( − ) + 11 = 20 . Thus 2 1 ≤ 1 . Since
4
4
2
5 5 5 5 5a −4a+11 20
2 2 2 2 4 1 < 1
5b − 4b + 11 ≥ 5 · ( ) − 4 · ( ) + 11 = 11 − > 10 , then 2 . Similarly, we
5 5 5 5b −4b+11 10
1
1
1
1
1
1
1
1
1
have 5c −4c+11 < 10 . Hence, 5a −4a+11 + 5b −4b+11 + 5c −4c+11 < 20 + 10 + 10 = .
2
2
2
2
4
2 .
n
2
3
1
3.91 Given a natural number n> 1, show C + C + C + ··· + C >n × 2 n−1
n
n
n
n
Proof: According to Binomial theorem, we have 2 = (1 + 1) = 1+ C + C + ··· + C , thus
n
1
n
n
2
n
n
n
C + C + C + ··· + C =2 − 1. On the other hand, the geometric series with first term 1
1
n
n
3
2
n
n
n
n
n
n
3
and common ratio 2 is S n = 1·(1−2 ) =2 − 1, i.e. 2 − 1 = 1+2+2 +2 + ··· +2 n−1 .
2
n
1−2
√
√
√
2
n
n
n
3 n−1
3
3
2
2 −1
2 −1 1+2+2 +2 +···+2
1+2+3+···+(n−1)
2 3
2
n
Therefore, = n = 2 1+2+2 +2 +···+2 n−1 √ > 1 × 2 × 2 × 2 × ··· × 2 n−1 n−1 n = 2 1+2+3+···+(n−1) = =
>
1 × 2 × 2 × 2 × ··· × 2 =
n n n n
n−1 n−1
2
n
n
1
3
n
n n(n−1) n(n−1) n−1 , that is, 2 − 1 >n × 2 2 . Hence, C + C + C + ··· + C >n × 2 2 .
n−1
2 2 2 =2 2 =2 2 n n n n
2
3.92 Positive numbers x,y,z satisfy x + y + z =1, find the minimum value
2
2
2
z
y
x
of 1−x 2 + 1−y 2 + 1−z 2.
Solution: (applying mean inequality)
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120

ELEMENTARY ALGEBRA EXERCISE BOOK I inequAlities





5
1
1
2
1
1
5
x + 3 3 2 2 2 5 5 3 3 1 2 2 3 3 1 2 2 3 3 x · 5 3 3 1 2 3 3 1 2 2 3 3
√ x = x .
√ x ≥ 3
√ x +
√ x = x +
5
√ x · 2
√ x = x +
√ x = x
√ x ≥ 3
√ x +
x +
x ·
√ x ·
3 3
3 3
3 3
3 3
3 3
2
2
5
3
2
3
2
5
Similarly, we can obtain y + √ y ≥ y ,z + √ z ≥ z .
3 3 3 3
2
2
5
5
3
3
2
2
5
3
Add these three inequalities up to obtain x + y + z + √ (x + y + z ) ≥ x + y + z .
3 3
2
3
3
5
5
3
5
2
2
2
Since x + y + z =1, then x + y + z + √ ≥ x + y + z ,
3 3
3 2 3 2 3 2 √ (i).
2
then x (1 − x )+ y (1 − y )+ z (1 − z ) ≤
3 3
y
2

3
3
z
2
2
3
3
2
y
x
3
[x (1 − x )+ y (1 − y )+ z (1 − z )]( 2 x 2 + 1−y + 2 + 1−z + 2 ) ≥ ( x (1 − x ) 2 1−x 2 + 2 +
3
z
3
x
2
2
x
3
2 ) ≥ (
[x (1 − x )+ y (1 − y )+ z (1 − z
1−x )](
x (1 − x )
1−x 2 1−y 2 1−z 1−x

y
2
2
2
3
y (1 − y ) 2 1−y 2 + 2 + z (1 − z ) 2 1−z 2 = x + y + z =1
3
z
2
2
y
z
2
2 = x + y + z =1 . Thus
2
2
3
3
z (1 − z )
y (1 − y )
1−y
1−z
x x 2 + + y y 2 + + z z 1 1
2 2
3
3 x (1−x )+y (1−y )+z (1−z
2
2 2
3 3
2 2
3 3
1−x
1−y
1−z
1−x 2 1−y 2 1−z 2 ≥ ≥ x (1−x )+y (1−y )+z (1−z ) ) (ii). From (i) and (ii), we obtain
√
x y z 3 3
1−x 2 + 1−y 2 + 1−z 2 ≥ 2 .
√
1 x y z 3 3
When x = y = z = √ , 2 + 2 + 2 reaches the minimum value .
3 1−x 1−y 1−z 2
3.93 Positive numbers a 1 ,a 2 , ··· ,a n and b 1 ,b 2 , ··· ,b n
satisfy a 1 + a 2 + ··· + a n ≤ 1,b 1 + b 2 + ··· + b n ≤ n ,
n
show ( 1 + 1 )( 1 + 1 ) ···( 1 + 1 ) ≥ (n + 1) .
a 1 b 1 a 2 b 2 a n b n
) ≤
Proof: The given conditions together with Mean Inequality result in a 1 a 2 ··· a n ≤ ( a 1 +a 2 +···+a n n 1
n n n

n


(i), and b 1 b 2 ··· b n ≤ ( b 1 +b 2+···+b n )=1 (ii). In addition, 1 + 1 = 1 + ··· + 1 + 1 ≥ (n + 1) n+1 1 1
n
a i b i na i na i b i na i b i

n

1 1 1 1 1 1 1 n terms
+ = + ··· + + ≥ (n + 1) n+1 (i =1, 2, ··· ,n ) (iii).
a i b i na i na i b i na i b i

n terms
From (i),(ii),(iii), we can obtain

n
1 1 1 1 1 1 1 1 1
+ + ··· + ≥ (n + 1) n n+1 · ·
n n
a 1 b 1 a 2 b 2 a n b n (n ) a 1 a 2 ··· a n b 1 b 2 ··· b n

1
n n
≥ (n + 1) n n+1 · (n ) · 1
n n
(n )
n
=(n + 1) .
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121