Is this simple-looking moment inequality true?

Let $p \ge 1$ be an integer. Does there exist a constant $C_p$ such that for every random variable $X \ge 0$, $$\mathbb{E} \left[ \left(X - \mathbb{E} \left[ X \right] \right)^{2p} \right] \le C_p \mathbb{E} \left[ \left(X^p - \mathbb{E} \left[ X^p \right] \right)^{2} \right] \ \ ?$$

• It would have been better, I think, to write your answer as an actual answer, not as an edit to the question, so that people can see that the question has been answered. That being said, it's a little strange to ask a question on this site and then provide the answer yourself only 19 hours later... – Tom De Medts May 13 '16 at 11:30
• Sorry, I fixed the edition problem. About the time lag, well, I had given it a serious try before posting, but had not thought about this symmetrization trick. – Elwood May 13 '16 at 16:19
• @TomDeMedts -- it is indeed strange, but I call this phenomenon "MO pressure" – sometimes after posting the question to which one has given sufficient thought, the mere fact of making it public puts pressure on the poster that can sometimes lead to progress or even an answer shortly after posting the question! – Suvrit May 13 '16 at 22:45

The answer is yes. Let $Y$ be an independent random variable distributed as $X$. We have $$\|X - \mathbb{E}[X]\|_{L^{2p}} = \|\mathbb{E}\left[X - Y \, | \, X\right]\|_{L^{2p}} \le \|X - Y\|_{L^{2p}}.$$ Moreover, there exists a constant $C_p$ such that for every $x,y \ge 0$, $$|x-y|^p \le C_p |x^p - y^p|.$$ Indeed, it suffices to verify this for $x = 1$ and $y \in [0,1]$ by homogeneity and symmetry. This is then a simple exercise. As a consequence, we deduce $$\|X - \mathbb{E}[X]\|_{L^{2p}} \le C_p \||X^p - Y^p|^{1/p}\|_{L^{2p}} = C_p \|X^p - Y^p\|_{L^{2}}^{1/p}.$$ We conclude by the triangle inequality.