An important inequality in Analysis
Let p, q be conjugate exponents. Let a, b be any positive real numbers. Prove that
ab ≤ ----- + -----
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.
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.
- myisland8132Lv 71 十年前最愛解答
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。
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:
Here we used the defining property of convex functions: for any t between 0 and 1 inclusively,