The map $g(t)=(E|X|^t)^{\frac 1t}$ is monotonic. 
Let $X$ be a random variable. Prove that $g:(0,\infty)\rightarrow[0,\infty]$ which is defined by $g(t)=(E|X|^t)^{\frac 1t}$ ($E$ marks the expected value) is monotonic.

I tried experimenting with Markov's inequality but I had a hard time dealing with it since $X$ is not in a specific class of random variables.
 A: Fix $s\lt t$ and define $p:=t/s \gt 1$. The map $x\mapsto |x|^p$, is convex, 
hence for each non-negative random variable $Y$, we have
$$\left(\mathbb E[Y]\right)^p\leqslant \mathbb E[Y^p].$$
Using this with $Y=|X|^s$, we derive that 
$$\left(\mathbb E[|X|^s]\right)^{t/s}  \leqslant \mathbb E[|X|^{s\cdot t/s} ],$$
from which the wanted inequality follows.
A: This follows from the general Lyapunov inequality which states that $$\nu_a^{b-c}\nu_c^{a-b}\ge\nu_b^{a-c}\quad\forall a>b>c\ge0,\quad\text{where}\quad \nu_t=\mathbb{E}\left(|X|^t\right)$$
Another form of this inequality is proven here using Hölder's inequality.
Choosing $c=0$ in the generalised inequality we have, 
$\nu_a^b>\nu_b^a\quad\forall a>b>0$
$\implies \nu_a^{1/a}>\nu_b^{1/b}\quad\forall a>b>0$
$\implies \nu_t^{1/t}$ is non-decreasing in $t\quad\forall t>0$.

Note that proving the generalised Lyapunov inequality is equivalent to proving $$\alpha g(a)+(1-\alpha)g(c)\ge g\left(\alpha a+(1-\alpha)c\right),$$ where $\alpha=\frac{b-c}{a-c}$ and $g(t)=\ln(\nu_t)$. 
Then showing $g$ is convex (by proving $g''\ge0$) proves the inequality.
