TITLE: Prove $\mathbf{E}[X^2] \geq \frac{ \mathbf{E}[X^{2\alpha}]}{ \mathbf{E} [X^{\alpha}]^2}$
QUESTION [7 upvotes]: I have a square-integrable random variable $X > 0$ with $\mathbf{E}[X] = 1$ and $0 < \alpha < 1$.
I would like to prove that:
\begin{equation}
\mathbf{E}[X^2] \geq \frac{ \mathbf{E}[X^{2\alpha}]}{ \mathbf{E} [X^{\alpha}]^2}
\end{equation}
It seems to be true numerically, but I can't come up with the analytical proof. Any ideas? Thank you!

REPLY [4 votes]: The required inequality follows from two applications of Hölder's inequality.  Firstly, note that
$$\Big(\mathbb{E}\left[X^2\right]\Big)^{\frac{1-\alpha}{2-\alpha}}\,\Big(\mathbb{E}\left[X^\alpha\right]\Big)^{\frac{1}{2-\alpha}}\geq \mathbb{E}\left[X\right]\,.$$
Therefore,
$$\frac{\mathbb{E}\left[X^2\right]}{\big(\mathbb{E}\left[X\right]\big)^2}\geq \frac{\Big(\mathbb{E}\left[X^2\right]\Big)^{\frac{\alpha}{2-\alpha}}}{\Big(\mathbb{E}\left[X^\alpha\right]\Big)^{\frac{2}{2-\alpha}}}\,.$$
Secondly, we have
$$\Big(\mathbb{E}\left[X^2\right]\Big)^{\frac{\alpha}{2-\alpha}}\,\Big(\mathbb{E}\left[X^\alpha\right]\Big)^{\frac{2-2\alpha}{2-\alpha}}\geq \mathbb{E}\left[X^{2\alpha}\right]\,.$$
This shows that
$$\frac{\Big(\mathbb{E}\left[X^2\right]\Big)^{\frac{\alpha}{2-\alpha}}}{\Big(\mathbb{E}\left[X^\alpha\right]\Big)^{\frac{2}{2-\alpha}}}\geq \frac{\mathbb{E}\left[X^{2\alpha}\right]}{\big(\mathbb{E}\left[X^{\alpha}\right]\big)^2}\,.$$
Thus, for any nonnegative random variable $X$ such that $\mathbb{E}\left[X^2\right]$ is a finite positive real number, we have
$$\frac{\mathbb{E}\left[X^2\right]}{\big(\mathbb{E}\left[X\right]\big)^2} \geq  \frac{\mathbb{E}\left[X^{2\alpha}\right]}{\big(\mathbb{E}\left[X^{\alpha}\right]\big)^2}$$
for any $\alpha\in [0,1]$ (where $0^0$ is defined to be $1$).  The inequality becomes an equality if and only if $\alpha=1$ or $X$ is almost surely constant.  
As before, $X$ is a nonnegative random variable such that $\mathbb{E}\left[X^2\right]$ is a finite positive real number.  Let $\beta\in[1,\infty]$ denote the supremum of the set of positive real numbers $\alpha$ such that $\mathbb{E}\left[X^{2\alpha}\right]$ is finite.  The required inequality also shows that the function $f:[0,\beta)\to[0,\infty)$ defined by
$$f(\alpha):=\frac{\mathbb{E}\left[X^{2\alpha}\right]}{\big(\mathbb{E}\left[X^{\alpha}\right]\big)^2}$$
for every $\alpha\in[0,\beta)$ is nondecreasing.  It is strictly increasing unless $X$ is almost surely constant.