Show that for real numbers $a,b\ge0$ $ab\le\frac{a^p}{p}+\frac{b^q}{q}$

Question

Let $p$ and $q$ be positive real numbers such that $\frac{1}{p}+\frac{1}{q}=1$. Show that for real numbers $a,b\ge0$

$ab\le\frac{a^p}{p}+\frac{b^q}{q}$

I do not know how to approach this. Any help/hint will be appreciated.

Duplicate: https://math.stackexchange.com/questions/357767/intuition-behind-a-b-p-q-subset-mathbbr-wedge-1-p-1-q and https://math.stackexchange.com/questions/676803/prove-that-xy-leq-fracxpp-fracyqq?noredirect=1&lq=1 — Mythomorphic, Aug 14 '17 at 07:46
You can find the proof here: https://en.wikipedia.org/wiki/Young%27s_inequality_for_products — ajotatxe, Aug 14 '17 at 07:49

score 1 · Accepted Answer · answered Aug 14 '17 at 07:50

1

I have posted a link with the proof in a comment. If you prefer a hint, use that the function $\ln$ is concave.

answered Aug 14 '17 at 07:50

ajotatxe

Many thanks! I wasn't aware that it is a well known inequality. – user467745 Aug 14 '17 at 07:54

1 Answers1