Page 120 - Elementary Algebra Exercise Book I

ELEMENTARY ALGEBRA EXERCISE BOOK I inequAlities

1
2
3.89 Let a 1 ,a 2 , ··· ,a n be a permutation of 1, 2, ··· ,n , show + + ··· + n−1 ≤ a 1 + a 2 + ··· + a n−1
2
n
3
a 3
a n
a 2
1 + + ··· + n−1 a 1 + a 2 + ··· + a n−1 .
2
2 3 n ≤ a 2 a 3 a n
Proof: Since a 1 ,a 2 , ··· ,a n is a permutation of 1, 2, ··· ,n , we have
(a 1 + 1)(a 2 + 1) · ··· · (a n−1 + 1) ≥ (1 + 1)(2 + 1) · ··· · (n − 1 + 1) = a 1 a 2 ··· a n a n . Thus
(a 1 + 1)(a 2 + 1) · ··· · (a n−1 + 1) ≥ (1 + 1)(2 + 1) · ··· · (n − 1 + 1) = a 1 a 2 ···
2
1a 1 +1
a 1 +1
1 a
1 + + ··· + n−1 a n−1 a 1 + + +···+ + a n−1n−1 1 a 1 1 1 a 2 1 a 1 a n−1n−1 111 1 1 + a 2 +1 1 1 1 a n−1 +1 + a 2 1 +1 1 1 + a 1 +1+1 a 2 +1 1 a n−1 +1 a n−1 +1
+ ···+
+ + +···++ +···++ + +···+
a n−1
+ + +···+
=
a n−1
a n−1
a 1
1a 2
a 2
a 1
1 a 2
=+···+ 1
+ ≥1
+···++ + +···+= 1
+···+ a 1 +1
+ 1
++ +···+ a n−1
≥ a n−1
=+ + +···++ a 1
a 2
2 3 a 2 + +···+≤ a n a 2 1 a 3 2a 3 n a n a a 12 1 a2 a n a nn + 2a+···+ a 3 a n a n + + +···+ a n 3 + + +···+ a 3 = + a n + ≥ a 2 +1 +···+ ≥
a
n
a
a 1 1a 1
a 23
a 3
a 1 a 1
a 22
a 2
a 2
a n a 2
a 1
a 2 a n
a n a n a 2 a 3 a n 1 2 a n a 1 a 2 a n a 1 a 2 a 3 a n

n (a 1 +1)(a 2 +1)·····(a n−1 +1)+1) a n−1 +1
a 2 +1
1
1
1
a 1 +1
1
n
+ +···+ + + +···+ n + + +···+ n= + n +···+ a 1 a 2 ···a n n n (a 1 +1)(a ≥ n . In
≥ n +
≥ n 2 +1)·····(a n−1 +1)
a 1 a 2 a n−1 1 1 a n−1 n (a 1 +1)(a 2 +1)·····(a n−1 +1)(a 1 +1)(a 2 +1)·····(a n−1 ≥ n
≥
a
a 2 a 3 a n 1 2 a n a 1 a 2 a 1 a 2 ···a na n a 1 a 2 a 1 a 2 ···a n 3 a n
a 1 a 2 ···a n
1
2
1
1
1
n n (a 1 +1)(a 2 +1)·····(a n−1 +1) ≥ n addition, n =( + + ··· + ) + ( + + ··· + n−1 ). Hence,
a 1 a 2 ···a n 1 2 n 2 3 n
a
11 + + ··· ++ + ··· + n−1n−1 a 1 a 1 ++ a 2 a 2 + ··· ++ ··· + a n−1n−1 .
22
2 2 3 3 n n ≤≤ a 2 a 2 a 3 a 3 a nn
a
3.90 If real numbers a, b, c satisfy a + b + c =3, show
1 + 1 + 1 ≤ .
1
2
2
2
5a −4a+11 5b −4b+11 5c −4c+11 4
9
Proof: If a, b, c are all less than , then we can show 2 1 ≤ 1 (3 − a) ( ). Actually
5 5a −4a+11 24 2 3 2 2
( ) ⇔ (3−a)(5a −4a+11) ≥ 24 ⇔ 5a −19a +23a−9 ≤ 0 ⇔ (a−1) (5a−9) ≤
2
2
2
3
( ) ⇔ (3−a)(5a −4a+11) ≥ 24 ⇔ 5a −19a +23a−9 ≤ 0 ⇔ (a−1) (5a−9) ≤ 0 ⇔ a< 9 5

3
2
2
2
( ) ⇔
0 ⇔ a< 9(3−a)(5a −4a+11) ≥ 24 ⇔ 5a −19a +23a−9 ≤ 0 ⇔ (a−1) (5a−9) ≤
5
0 ⇔ a< . Similarly, we can obtain 2 1 ≤ 1 (3 − b), 2 1 ≤ 1 (3 − c) . Add these
9
5 5b −4b+11 24 5c −4c+11 24
1
1
1
1
1
1
1
three inequalities up to obtain 1 5a −4a+11 1 5b −4b+11 1 5c −4c+11 1 ≤ 24 (3−a)+ (3−b)+ (3−c) = 24 [9−(a+b+c)] =
+
+

2
2
2
1
24
1
1
24
2
1 2
24
24
5a −4a+11 + 5b 2 1 −4b+11 + 5c −4c+11 ≤ 24 (3−a)+ (3−b)+ (3−c) = 24 [9−(a+b+c)] =
[9 − 3] =
1
1 + 1 + 1 1 (3−a)+ (3−b)+ (3−c) = 1 [9−(a+b+c)] = 1 [9 − 3] = 1 . 4
1
24
2
2
2
5a −4a+11 5b −4b+11 5c −4c+11 ≤ 24 24 24 24 24 4
1 [9 − 3] = 1
24 4
9
9
If at least one of a, b, c is not less than , without loss of generality, assume a ≥ , then
5 5
9
9
5a − 4a + 11 = 5a(a − ) + 11 ≥ 5 · · ( − ) + 11 = 20 . Thus 2 1 ≤ 1 . Since
4
4
2
5 5 5 5 5a −4a+11 20
2 2 2 2 4 1 < 1
5b − 4b + 11 ≥ 5 · ( ) − 4 · ( ) + 11 = 11 − > 10 , then 2 . Similarly, we
5 5 5 5b −4b+11 10
1
1
1
1
1
1
1
1
1
have 5c −4c+11 < 10 . Hence, 5a −4a+11 + 5b −4b+11 + 5c −4c+11 < 20 + 10 + 10 = .
2
2
2
2
4
2 .
n
2
3
1
3.91 Given a natural number n> 1, show C + C + C + ··· + C >n × 2 n−1
n
n
n
n
Proof: According to Binomial theorem, we have 2 = (1 + 1) = 1+ C + C + ··· + C , thus
n
1
n
n
2
n
n
n
C + C + C + ··· + C =2 − 1. On the other hand, the geometric series with first term 1
1
n
n
3
2
n
n
n
n
n
n
3
and common ratio 2 is S n = 1·(1−2 ) =2 − 1, i.e. 2 − 1 = 1+2+2 +2 + ··· +2 n−1 .
2
n
1−2
√
√
√
2
n
n
n
3 n−1
3
3
2
2 −1
2 −1 1+2+2 +2 +···+2
1+2+3+···+(n−1)
2 3
2
n
Therefore, = n = 2 1+2+2 +2 +···+2 n−1 √ > 1 × 2 × 2 × 2 × ··· × 2 n−1 n−1 n = 2 1+2+3+···+(n−1) = =
>
1 × 2 × 2 × 2 × ··· × 2 =
n n n n
n−1 n−1
2
n
n
1
3
n
n n(n−1) n(n−1) n−1 , that is, 2 − 1 >n × 2 2 . Hence, C + C + C + ··· + C >n × 2 2 .
n−1
2 2 2 =2 2 =2 2 n n n n
2
3.92 Positive numbers x,y,z satisfy x + y + z =1, find the minimum value
2
2
2
z
y
x
of 1−x 2 + 1−y 2 + 1−z 2.
Solution: (applying mean inequality)
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