An important inequality in Analysis

2007-07-11 4:04 pm

Question:
Let p, q be conjugate exponents. Let a, b be any positive real numbers. Prove that
a^p b^q
ab ≤ ----- + -----
 p q
This is trivially true for a = b = 0. Please give only the answer for positive part.

* I won't give any marks to someone who is giving pictorial answers, I will only give marks to whom can give the analytical proof.

更新1:

Definition: Let p > 1, define q by 1 / p + 1 / q = 1, then p and q are called " conjugate exponents". * It has good properties, e.g., (p - 1)(q - 1) = 1, but I am not asking the proof of this.

回答 (1)

myisland8132

2007-07-11 5:09 pm

✔ 最佳答案

這條即是Young's inequality 。其中一個證明可以在
A comprehensive course in pure mathematics : Algebra I CS Lee, Learner's Publishing Co. Ltd
Chapter 5 找到﹐裡面只用了中學程度的微積分﹐不過證明較長。
另一個證明可以在維基找到﹐用了凸函數的性質﹐簡明快捷。
Proof
The proof is trivial if a = 0 and/or b = 0. Therefore, assume a,b > 0.
The function f(x) = ex is convex, since its second derivative is positive for any value. Thus, it follows:

圖片參考：http://upload.wikimedia.org/math/a/9/7/a970f131ff86a630e1d53df5d062f0bf.png

Here we used the defining property of convex functions: for any t between 0 and 1 inclusively,

圖片參考：http://upload.wikimedia.org/math/8/3/2/8321d5e35870963c4aab00502f63ff60.png

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