Prove that $xy \leq\frac{x^p}{p} + \frac{y^q}{q}$
Solution 1:
The exponential funtion $t\mapsto \exp(t)$ is convex, so $$\begin{align} xy&=\exp(\log(xy))\\ &=\exp(\log(x)+\log(y))\\ &=\exp((1/p)\log(x^p)+(1/q)\log(y^q))\\ &\leq (1/p)\exp(\log(x^p))+(1/q)\exp(\log(y^q))\\ &=\frac{x^p}{p}+\frac{y^q}{q}\\ \end{align}$$
Solution 2:
It is the so-called Young's inequality and to prove it you can exploit the concavity of the logarithm: $$ \log (xy) = \frac{1}{p} \log x^p + \frac{1}{q} \log y^q \le \log \left( \frac{1}{p} x^p + \frac{1}{q} y^q \right). $$