Page 117 - Elementary Algebra Exercise Book I

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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