TITLE: Show $(\mathbb{E}\vert X^2 - \mathbb{E}[X^2]\vert)^2 \leq 4\mathbb{E}[X^2](\mathbb{E}[X^2]-\mathbb{E}[X]^2)$
QUESTION [5 upvotes]: Anyone have any leads on this? X has a finite second moment and is nonnegative.
\begin{equation}
(\mathbb{E}\vert X^2 - \mathbb{E}[X^2]\vert)^2 \leq 4\mathbb{E}[X^2](\mathbb{E}[X^2]-\mathbb{E}[X]^2)
\end{equation}
Here's what I have:
$\mathbb{E}[4X^2]^{1/2}[\mathbb{E}[(X-\mathbb{E}[X])^2]^{1/2}\geq \mathbb{E}\vert 2X(X-\mathbb{E}[X])\vert = \mathbb{E}\vert 2X^2 - X\mathbb{E}[X]\vert $
By Cauchy-Schwarz, and then squaring both sides we get:
$[\mathbb{E}\vert 2X^2 - X\mathbb{E}[X]\vert]^2\leq \mathbb{E}[4X^2][\mathbb{E}[(X-\mathbb{E}[X]^2]$
But I cant figure out what to do next or if this is even in the right direction.

REPLY [2 votes]: You could start with:
\begin{eqnarray*}
\left(E\vert X^2 - E[X^2]\vert\right)^2 &=& \left(E\left| X - \sqrt{E[X^2]}\right| \left| X + \sqrt{E[X^2]}\right|\right)^2 \\
&\leq& E\left(\left(X-\sqrt{E(X^2)}\right)^2\right) E\left(\left(X+\sqrt{E(X^2)}\right)^2\right) \qquad\text{by Cauchy-Schwarz.}
\end{eqnarray*}
Then expand both squares and use linearity of expectation.