AM GM inequality using a relation

Question

I am given a relation that for $a,b>0$ we have $$ab\leq \dfrac{a^p}{p}+\dfrac{b^q}{q}$$ where $ p$ and $ q$ satisfy $\dfrac{1}{p}+\dfrac{1}{q}=1$. Now how do I use this to prove the AM-GM inequality?

Answer 1

If you introduce $x=a^p$, $y=b^q$ $s=1/p$ and $t=1/q$, then this is the weighted AM-GM: $$ x^sy^t\leq sx+ty $$ You can deduce this from the regular AM-GM for rational $s$ and $t$, but for real $s$ and $t$ you need some additional argument, like continuity.