Unknown

358 Answers to Exercises and Solutions to Problems

3.5. (a) Wi = l/n.

(c)

WI TJ..

a1 a2 .awn= n a1 w1... a~~~2(a~-lu~)W*-1+W*

5 wlal + ... + wn-2an-2 + (w,-I + w,-2)0~-,0~

5 wlal +. .. + Wn---2h-2 + (~~-1 + wu,-2)(uan-1 + vu,),

which yields the result by (b).

3.7. (a) Apply the AGM inequality to x and x-l.

(b) 1 - 4x(1 - x) = (2x - 1)2 > 0.

3.8. This is a consequence of the CSB inequality applied to u:12, uiJ2,...

and .;iJ2, a;“‘,....

3.9. (a) When n = 1, equality holds without any restriction on x1. Let

n = 2 and suppose only that both variables have the same sign. If either

vanishes, we have essentially the n = 1 case. Otherwise,

(1+ x1)(1 + x2) = 1 +x1 + x2 + 21x2 > 1+ z1+ X2.

For n 2 3, there is a simple induction argument for the case that all

xi 2 - 1. Assuming the result holds for n - 1, we find that the left side

exceeds (1 + xi + 22 +... + x,,-i)(l + xn) which in turn exceeds the right

side by the n = 2 case. (Where is the condition xi 2 -1 used?)

However, with the stronger hypothesis on the xi, a more delicate argu-

ment is needed. Again, assume the result for the n - 1 case. Then (assuming

for convenience all Xi are nonzero),

= [(1+xl)~~~(1+x,~l)-(1+x1+~~~+x~-l)]

+xc,[(l+xl)***(l+x,-1)-l].

The first term on the right is positive by the induction hypothesis. If all

the ti are positive, then the second term is clearly positive. If all the zi are

negative, then -1 5 1 + xi 5 1, and both factors of the second term are

negative. If any of the xi vanishes, we essentially have the case of a lower

n. Equality holds iff at most one of the xi is nonzero.

Solutions to Problems

Chapter 7

4.1. By the CSB inequality applied to the quartuples (1 - x), (x - y),
(y - z), % and 1, 1, 1, 1, we have that

1 = (l-x)+(x--y)+(y-z)+z

5 41/2[(1 - x)” + (x - y)2 + (y - %)” + %2]i’2 = 1.