Purely “algebraic” proof of Young’s Inequality

Young’s inequality states that if a,b0, p,q>0, and 1p+1q=1, then abapp+bqq (with equality only when ap=bq). Back when I was in my first course in real analysis, I was assigned this as homework, but I couldn’t figure it out. I kept trying to manipulate the expressions algebraically, and I couldn’t get anywhere. But every proof that I’ve seen since uses calculus in some way to prove this. For example, a common proof is based on this proof without words and integration. The proof on Wikipedia uses the fact that log is concave, which I believe requires the analytic definition of the logarithm to prove (correct me if I’m wrong).

Can this be proven using just algebraic manipulations? I know that that is a somewhat vague question, because “algebraic” is not well-defined, but I’m not sure how to make it more rigorous. But for example, the proof when p=q=2 is something I would consider to be “purely algebraic”:

0(ab)2=a2+b22ab, so aba22+b22.

Answer

This proof is from “Mathematical Toolchest” published by the Australian Mathematics Trust (image).

Example. If p and q are positive rationals such that 1p+1q=1, then for positive x and y xpp+yqqxy.

Since 1p+1q=1, we can write p=m+nm, q=m+nn where m and n are positive integers. Write x=a1/p, y=b1/q. Then xpp+yqq=am+nm+bm+nn=ma+nbm+n.

However, by the AM–GM inequality, ma+nbm+n(ambn)1m+n=a1pb1q=xy, and thus xpp+yqqxy.

Attribution
Source : Link , Question Author : asmeurer , Answer Author : Community

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