Page 90 - Elementary Algebra Exercise Book I

ELEMENTARY ALGEBRA EXERCISE BOOK I inequAlities

3 INEQUALITIES

3.1 Determine the order of the numbers , log 2, .
4
2
9 5 5
4
4
4
Solution: − log 2= 4 − lg 2 = 4 lg 5−9 lg 2 = lg 5 −lg 2 9 = lg 625−lg 512 > 0, thus > log 2.
9 5 9 lg 5 9lg5 9 lg 5 9 lg 5 9 5
2
2
4
2 lg 2 2 5 lg 2−2 lg 5 lg 32−lg 25 , thus log 2 > . Hence, > log 2 > .
log 2 − 5 = lg 5 − 5 = 5lg5 = 5 lg 5 > 0 5 5 9 5 5
5
3.2 Solve the inequality 1 < log 2 x< 1.
100 0.1
√
Solution: 1 < log 2 x< 1 ⇒ 1 < log x< 1 or −1 < log x< − 1 ⇒ 0.1 <x< 10 0.1
100 0.1 0.1 10 0.1 10
√
or 10 10 <x< 10.
√ √
3.3 a, b, c, d are positive numbers, show (a + c)(b + d) ≥ ab + cd .
√
√
√
√
Proof: ad + bc ≥ 2 abcd ⇔ ad + bc + ab + cd ≥ ab +2 abcd + cd ⇔ (a + c)(b + d) ≥
ad
√
√
√
√ √+ bc ≥ 2 abcd ⇔ ad + bc + ab + cd ≥ ab +2 abcd + cd ⇔ (a + c)(b + d) ≥

√
√
√
2

( ab +
ab +
2
( ab + cd) ⇔ (a + c)(b + d) ≥ ab + cd cd since a, b, c, d > 0.
(a + c)(b + d) ≥
cd) ⇔
3.4 Given −1 ≤ u + v ≤ 1, 1 ≤ u − 2v ≤ 3, find the range of 2u +5v .
Solution: Let 2u +5v = λ 1 (u + v)+ λ 2 (u − 2v)=(λ 1 + λ 2 )u +(λ 1 − 2λ 2 )v2u +5v = λ 1 (u + v)+ λ 2 (u − 2v)=(λ 1 + λ 2 )u +(λ 1 − 2λ 2 )v ,

then λ 1 + λ 2 =2 and λ 1 − 2λ 2 =5. Solve them to obtain λ 1 =3,λ 2 = −1.
Hence, 2u +5v = 3(u + v) − 1(u − 2v) ∈ [(−1) × 3 − 3, 1 × 3 − 1] = [−6, 2]2u +5v = 3(u + v) − 1(u − 2v) ∈ [(−1) × 3 − 3, 1 × 3 − 1] = [−6, 2].

3.5 Given |h| < , |k| < , show |2h − 3k| <ε .
ε
ε
4 6
ε
ε
ε
Proof: |h| < ε ⇔− <h < ε ⇔− < 2h< .
4 4 4 2 2
ε
ε
ε
|k| < ε ⇔− <k < ε ⇔− < 3k< .
6 6 6 2 2
Hence, −ε< 2h − 3k <ε ⇔|2h − 3k| <ε .
1 3 5 99 1
3.6 Show the inequality · · · ··· · < .
2 4 6 100 10
1
2 3
4
3
4
1
Proof: < , < , ··· , 97 < 98 , 99 < 100 . Multiply all these inequalities: · · ··· · 97 · 99 < 2 · · ··· · 98 · 100
2 3 4 5 98 99 100 101 2 4 98 100 3 5 99 101
3
4
1 · · ··· · 97 · 99 < 2 · · ··· · 98 · 100 . Multiply this inequality by · · ··· · 97 · 99 : ( · · ··· · 97 · 99 2 1 .
3
1
1
3
) <
2 4 98 100 3 5 99 101 2 4 98 100 2 4 98 100 101
1 3 97 99 1 < 1
Take the square root to obtain 2 · · ··· · 98 · 100 < √ 101 10 .
4
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