Let $X$ be a random variable and let $p\in(0,\infty)$ such that $\mathbf{E}(|X|^p)<\infty$. Prove that for all $q\in(0,p)$, we have $\mathbf{E}(|X|^q)\leqslant [\mathbf{E}(|X|^p)]^{\frac{q}{p}}$. You can use that $|x|^r$ is convex for all $r\geqslant 1$.
My try. Consider the random variable $X^p$. From $p>0$ and $0<q<p$, we have $0<\frac{q}{p}<1$. This means that $f(x)= |x|^{q/p}$ is concave, but then $g(x)= -|x|^{q/p}$ is convex. From Jensen's Inequality we get $\mathbf{E}[-f(X^p)]=-\mathbf{E}[f(X^p)]\geqslant -f[\mathbf{E}(X^p)] \iff \mathbf{E}[f(X^p)]\leqslant f[\mathbf{E}(X^p)]$. This gives $$\mathbf{E}[|X^p|^{\frac{q}{p}}]\leqslant [\mathbf{E}(|X^p|)]^{\frac{q}{p}} \iff \mathbf{E}(|X|^q)\leqslant [\mathbf{E}(|X|^p)]^{\frac{q}{p}}$$ as asked.
My thoughts
I would like to know how I can use the fact that $\mathbf{E}(|X|^p)<\infty$. Also, when I look at the graph of $g$, it is not convex for all values of $x$, only on the right and left branch separately. Do I need to split cases in my proof for that? I would also like to know if there is a cleaner or better proof.
You may look at $g(x)=-x^{q/p}$ on $x\geq 0$ side, and consider the random variable $|X|^{p}$. The assumption that $E(|X|^{p})<\infty$ is for the purpose when you put the real number $E(|X|^{p})$ into $g(\cdot)$.
Another way:
For the convex function $\varphi(x)=x^{p/q}$ on $x\geq 0$, then $[E(|X|^{q})]^{p/q}\leq E[\varphi(|X|^{q})]=E(|X|^{p})$, then $E(|X|^{q})\leq [E(|X|^{p})]^{q/p}$.