精華區beta Math 關於我們 聯絡資訊
※ 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. -- ※ 發信站: 批踢踢實業坊(ptt.cc) ◆ From: 140.112.218.142