Re: [分析] 初微(16)
※ 引述《PttFund (批踢踢基金)》之銘言：
: Let p, q > 1 with 1/p + 1/q = 1. Show that
: ab ≦ a^p/p + b^q/q
: where a and b are nonnegative.
: Note: 這個就是 Holder inequality, 然後請不要用 Young's inequality
: 來證明.
Another proof:
It is equivalent to show that for a,b≧0, 0 < λ < 1,
a^λ b^(1-λ) ≦ λa + (1-λ)b.
The result is obvious if b = 0; otherwise, dividing both
sides by b and setting t = a/b, we are reduced to showing
that
t^λ ≦ λt + (1-λ).
But by elementary calculus, t^λ-λt is strictly increasing
for t < 1 and strictly decreasing for t > 1, so its maximum
value, namely 1-λ, occurs at t = 1.
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