Page 104 - Elementary Algebra Exercise Book I

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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