$E[X^4]+E[X]^4 \geq 2E[X^2]^2$ for $X \geq 0$

Question

I was wondering whether this inequality is actually true. It seems to hold for discrete distributions, for the Normal and the Poisson distribution. Furthermore, after applying AM-GM, it can be seen that a sufficient condition for this inequality to hold is the following inequality $$ \|X\|_4 \|X\|_1 \geq \|X\|_2^2 $$ where the norms are the respective norms in $L_p$.

Answer 1

No. Consider $X$ following a Bernoulli distribution with parameter $p\in[0,1]$. Then, since $X^4 =X^2 = X$, your inequality is equivalent to $\mathbb{E}[X]+\mathbb{E}[X]^4 \geq 2\mathbb{E}[X]^2$, i.e., $$ p + p^4 \geq 2p^2 $$ which is not always true for $p\in[0,1]$. (E.g., it fails for $p=3/4$).