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An important inequality in Analysis

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:

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.

1 個解答

評分
  • 1 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。

    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/a970f131ff8...

    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/8/3/2/8321d5e3587...

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