Page 96 - Elementary Algebra Exercise Book I

ELEMENTARY ALGEBRA EXERCISE BOOK I inequAlities

√
4
3.24 Solve the inequality x log x > x · x (a> 0,a =1).
a
a 2
Solution: When a> 1, take the log with base a on both sides to obtain
(log x) > log x−2 ⇒ 2(log x) −9 log x+4 > 0 ⇒ (2 log x−1)(log x−4) >
2
2
9
2
9
2
(log x) 2 > log x−2 ⇒ 2(log x) −9 log x+4 > 0 ⇒ (2 log x−1)(log x−4) >
a
a
a
a
a
a
a
a
a
a
a
a
0 ⇒ log x< 1 1 2 √ 4
0 ⇒ log a 2 x< 2 or log x> 4 ⇒ 0 < x< a or x> a . When 0 <a < 1,
a
a
2 9 1 4
1
9
(log x) < 2 2 < log x − 2 ⇒ (2 log x − 1)(log x − 4) < 0 ⇒ 2 ⇒ < log x< 4 ⇒ a < 4 <
< log x< 4 ⇒ a
a
log x − 2 ⇒ (2 log x − 1)(log x − 4) < 0
a
a
a
√(log x)
a
x< a √ a 2 a a a 2 a
x< a .
1
3.25 Given a> 0,b > 0,c > 0,a + b + c =1, show (1 + )(1 + )(1 + ) ≥ 64.
1
1
a b c

√ √ √
Proof: Since a, b, c > 0,a + b + c =1 , then 1+ 1 a = 1+ a+b+c = 2+ b+c ≥ 2+ 2 bc ≥ 2 2 · 2 bc =4 a bc

a
a
a
a

√ √ √ 1 √ ca 1 √ ab
1+ 1 = 1+ a+b+c = 2+ b+c ≥ 2+ 2 bc ≥ 2 2 · 2 bc =4 bc . Similarly we can obtain 1+ ≥ 4 , 1+ ≥ 4 . Multiply
a a a a a a b b c c
√ √ √
1
1
1
these three inequalities to obtain (1 + )(1 + )(1 + ) ≥ 4 bc · 4 ca · 4 ab = 64.
a b c a b c
1
1
1
3.26 Show + n+1 + n+2 + ··· + n 1 2 > 1 for n ∈ N ,n ≥ 2.
n
1
1
1
Proof: + n+1 + n+2 + ··· + n 1 2 > n n 2 + n 1 2 + n 1 2 + ··· + n 1 2 = n+n(n−1) =1 .
2
n
n
√
√
3.27 Given x ≥ 0,y ≥ 0, show (x + y) + (x + y) ≥ x y + y x .
1
2
1
2 4
1
1
1
1
1
1
1
2 1
1
1
1
1
2
Proof: (x + y) + (x + y)=(x + y)[(x + y) + ]= 1 1 (x + y)[(x + ) +(y + )] ≥
(x + y) + (x + y)=
(x + y)[(x + y) + ]=(x + y)[(x + ) +(y + )] ≥
2 2 4 4 2 2 2 2 2 4 4 4 4
2
√ √ 1 1 1 1 √ √ 1 1 1 1 √ √ √ √ √ √ √ √ √ √
xy( x +y) = x y + y x .
x +2 y]=
xy[(x + ) +(y + )] ≥xy[2
xy[(x + ) +(y + )] ≥ xy[2 x +2 y]=xy( x + y) = x y + y x
4 4 4 4 4 4 4 4
2
1 x −x+1
3.28 Show ≤ x +x+1 ≤ 3.
2
3
2
2
2
2
Proof: Let y = x −x+1 , then yx + yx + y − x + x − 1 =0 ⇔ (y − 1)x +(y + 1)x + y − 1 =0

2
x +x+1
2 2 2 2 2 2 2
yx + yx + y − x + x − 1 =0 ⇔ (y − 1)x +(y + 1)x + y − 1 =0. Consider the discriminant Δ=(y+1) −4(y−1) = −3y +10y−3 ≥ 0 ⇒ 3y −10y+3 ≤ 0 ⇒ (3y−1)(y−3) ≤
2
2
2
2
Δ=(y+1) −4(y−1) = −3y +10y−3 ≥ 0 ⇒ 3y −10y+3 ≤ 0 ⇒ (3y−1)(y−3) ≤
≤ y ≤ 3
2 2 2 2 0 ⇒ 1 1 3 ≤ y ≤ 3.

Δ=(y+1) −4(y−1) = −3y +10y−3 ≥ 0 ⇒ 3y −10y+3 ≤ 0 ⇒ (3y−1)(y−3) ≤ 0 ⇒ 3
1
0 ⇒ ≤ y ≤ 3
3 a b
3.29 a, b, x, y are positive numbers and satisfy a + b = 10, + y =1, and x + y has
x
the minimum value 18, find the values of a, b .
√
b
a
Solution: The conditions imply that x+y =( + )(x+y)= a+b+ ay + bx = 10+ ay + bx ≥ 10+2 ay · bx = 10+2 ab

√ x y x y √ x y √ x y
b
a
x+y =( + )(x+y)= a+b+ ay + bx = 10+ ay + bx ≥ 10+2 ay · bx = 10+2 ab. Since x + y has the minimum value 18, then 10 + 2 ab = 18 ⇒ ab =4 ⇒ ab = 16
x y x y x y x y
√ √
10 + 2 ab = 18 ⇒ ab =4 ⇒ ab = 16. Solve
a + b = 10
ab = 16
to obtain a =2,b =8 or a =8,b =2.
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