If $E(X^2) = 1$ and $E(X^4)$ is finite then $E\vert{X}\vert \ge 1/\sqrt{E(X^4)}$

Question

I was trying the following problem :

Let $X$ be a random variable. Show that if $E\left(X^2\right) = 1$ and $E(X^4)$ is finite then $E\vert{X}\vert \ge 1/\sqrt{E\left(X^4\right)}$.

I have approached it by breaking up all the moments into the terms of central moments but it didn't work out.

Answer 1

Rewrite the inequality to be proved as $$E(X^2)^3\leqslant E(|X|)^2E(X^4)$$ or, equivalently, for some almost surely nonnegative $Z$ that I shall let you write down, $$E(Z^2)=E(Z^{2/3}\cdot Z^{4/3})=E(U\cdot V)\leqslant E(U^{3/2})^{2/3}E(V^3)^{1/3}$$ with $$(U,V)=(Z^{2/3},Z^{4/3})$$ and remember that a famous inequality states that, for every random variables $(U,V)$, $$E(U\cdot V)\leqslant E(U^p)^{1/p}E(V^q)^{1/q}$$ for every positive $(p,q)$ such that $$\frac1p+\frac1q=1$$

Answer 2

Apply the following generalized version of Holder's inequality $$E(U)E(V)E(W)\ge E((UVW)^{1/3})^3$$ valid for nonnegative random variables to $U=V=|X|$ and $W=X^4$.