Young's inequality

Theorem (Young’s inequality) Suppose $1<p<\infty$, then for all $a,b\ge 0$

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

where $\frac{1}{p}+\frac{1}{p^{\prime }}=1$.

Proof If either $a$ or $b$ equals zero then the inquality is trivial. Suppose that neither are zero, fix $b>0$ and define the real function:

$$f(a)=\frac{a^p}{p}+\frac{b^{p^{\prime }}}{p^{\prime }}-ab$$

its derivative is $f^{\prime }(a)=a^{p-1}-b$ so that $f$ is decrising in the inverval $(0,b^{1/p-1})$ and increasing in $(b^{1/p-1},\infty )$ so that it achieves a global minimum in $a=b^{1/p-1}$ with value $0$, so that

$$\frac{a^p}{p}+\frac{b^{p^{\prime }}}{p^{\prime }}-ab\ge 0\square$$