Proof of Minkowski's inequality

About Posts which Tagged by 'Probability'

Minkowski's inequality.  Let $X$ and $Y$ are random variables.  Let $1<p<\infty$. $$\left(\mathscr{E}|X+Y|^p\right)^{\frac{1}{p}}\leq \left(\mathscr{E}|X|^p\right)^{\frac{1}{p}}+\left(\mathscr{E}|Y|^p\right)^{\frac{1}{p}}.$$
See List of Inequalities.

$\bullet$ Proof.

$$\begin{array}{rl}\mathscr{E}|X+Y|^p &=\mathscr{E}\left(|X+Y||X+Y|^{p-1}\right) \\
(\mbox{Triangular Inequ.}\rightarrow)&\leq\mathscr{E}\left(|X||X+Y|^{p-1}\right)+\mathscr{E}\left(|Y||X+Y|^{p-1}\right) \\
(\mbox{H$\ddot{o}$lder's Inequ.}\rightarrow)&\leq\left(\mathscr{E}|X|^p\right)^{\frac{1}{p}}\left(\mathscr{E}|X+Y|^{q(p-1)}\right)^{\frac{1}{q}}+\left(\mathscr{E}|Y|^p\right)^{\frac{1}{p}}\left(\mathscr{E}|X+Y|^{q(p-1)}\right)^{\frac{1}{q}}\end{array}$$for $q$ such that $\frac{1}{p}+\frac{1}{q}=1$.  The condition on $p$ and $q$ implies that $q(p-1)=p$ and $\frac{1}{q}=1-\frac{1}{p}$, then $$\mathscr{E}|X+Y|^p\leq\left(\mathscr{E}|X|^p\right)^{\frac{1}{p}}\left(\mathscr{E}|X+Y|^p\right)^{1-\frac{1}{p}}+\left(\mathscr{E}|Y|^p\right)^{\frac{1}{p}}\left(\mathscr{E}|X+Y|^p\right)^{1-\frac{1}{p}}.$$Divide both side of the inequality by $\left(\mathscr{E}|X+Y|^p\right)^{1-\frac{1}{p}}$ then we have $$\left(\mathscr{E}|X+Y|^p\right)^{\frac{1}{p}}\leq \left(\mathscr{E}|X|^p\right)^{\frac{1}{p}}+\left(\mathscr{E}|Y|^p\right)^{\frac{1}{p}}.$$

$\Box$