Page 99 - Elementary Algebra Exercise Book I

ELEMENTARY ALGEBRA EXERCISE BOOK I inequAlities

3.35 Given |a| < 1, |b| < 1, |c| < 1 , show (1) |1 − abc| > |ab − c|; (2) a + b + c<abc +2 .

2
2 2
Proof: (1) The given conditions imply 1 − a b > 0, 1 − c > 0. Multiply them together to
obtain 1+ a b c >a b + c ⇒ 1 − 2abc + a b c >a b − 2abc + c ⇒ (1 − abc) >
2
2
2 2 2
2 2 2
2 2
2 2
2 2 2
2
2
2
2
2 2 2
2 2
2 2
1+ a b c >a b + c ⇒ 1 − 2abc + a b c >a b − 2abc + c ⇒ (1 − abc) >
(ab − c) ⇒|1 − abc| > |ab − c|.
2
2
(ab − c) ⇒|1 − abc| > |ab − c|
(2) (a − 1)(b − 1) > 0 ⇒ a + b < ab +1 (i). (ab − 1)(c − 1) > 0 ⇒ ab + c < abc +1 (ii).
(i)+(ii)⇒ a + b + c < abc +2.
2
3.36 The smaller root of the quadratic equation x − 5x log k + 6 log k =0 is in the
2
8
8
interval (1, 2). Find the range of the parameter k.
Solution: The parabola opens upward, and the smaller root is within (1, 2), then
f(1) > 0
f(2) < 0
log k> 1/2 or log k< 1/3
⇒ 8 8 ⇒ 2/3 < log k< 1 ⇒ 4 <k < 8.
2/3 < log k< 1 8
8
3.37 The inequality ax + bx + c> 0 has the solution set {x|α <x < β} where
2
0 <α <β . Find the solution set of the inequality cx + bx + a< 0.
2
⎧
⎨ α + β = −b/a > 0
Solution: The given condition implies that αβ = c/a > 0 and let the quadratic
a< 0
⎩
equation

1
b
cx + bx + a =0 has two roots x 1 ,x 2. Then x 1 + x 2 = − = α+β = 1 + ;
2
αβ
α
β
c
1
1
2
x 1 x 2 = a = 1 = 1 · . 0 <α <β ⇒ 1 < , in addition c< 0, then cx + bx + a< 0
c αβ α β β α
1
has the solution set {x|x< β 1 or x> }.
α
3.38 Real numbers a, b, x, y satisfy a + b =1,x + y =1, show |ax + by|≤ 1.
2
2
2
2
Proof 1: (|a|−|x|) ≥ 0 ⇒ a + x ≥ 2|ax|. Similarly we have b + y ≥ 2|by| .
2
2
2
2
2
Therefore, a + b + x + y ≥ 2(|ax| + |by|). Since a + b =1,x + y =1,
2
2
2
2
2
2
2
2
then 2 ≥ 2(|ax| + |by|) ≥ 2|ax + by| , thus |ax + by|≤ 1.
Proof 2: Since a + b =1,x + y =1, let a = sin θ, b = cos θ, x = sin ϕ, y = cos ϕ , then
2
2
2
2
ax + by = sin θ sin ϕ + cos θ cos ϕ = cos(θ − ϕ). Thus |ax + by| = | cos(θ − ϕ)|≤ 1.
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