How can I use the Cauchy-Schwarz inequality in this function of random variables? I have the function $\rho_{\lambda}:RV(\Omega)\rightarrow \mathbb{R}$ defined on the space $RV(\Omega)$ supported over some scenario set $\Omega$:
$\rho_\gamma(X)=\frac{1}{\gamma} \log (\mathbb{E}[e^{-\gamma X}])$
where $\gamma>0$. Now in my book they claim that the Cauchy-Schwarz inequality shows that $\rho_\gamma(2X)\geqslant2\rho_\gamma(X)$ for every random variable $X$ and every positive $\gamma$. I am having trouble seeing why this is the case though. Anyone have any ideas?
 A: 
Sorry I made a mistake. This is only a proof for the claim, and nothing is about Cauchy-Schwarz inequality.
Oh I am wrong again: Due to
@TheOscillator, Cauchy-Schwarz inequality is a way to prove $\mathbb{E}^2(X) \leq \mathbb{E}(X^2)$. (I usually prove it by $\mathbb{E}(X - \mathbb{E}X)^2 \geq 0$.)

The Cauchy-Schwarz inequality for random variables looks like this:
$$
\mathbb{E}(XY) \leq \sqrt{\mathbb{E}(X^2) \times \mathbb{E}(Y^2)}.
$$
If $Y \equiv 1$, then it becomes $\mathbb{E}^2(X) \leq \mathbb{E}(X^2)$, where $\mathbb{E}^2 X$ denotes $\left( \mathbb{E} X \right)^2$.
Now look at what we want to prove. After some reduction (cancelling out $\gamma$, removing the $\log$), you will find the claim turns into
$$
\mathbb{E}(e^{-2\gamma X}) \geq \mathbb{E}^2 (e^{-\gamma X}).
$$
$e^{-\gamma X}$ is just the “$X$” in Cauchy-Schwarz inequality.
A: By the way, you don't need Cauchy-Schwarz to prove $\operatorname E [X]^2 \le E[X^2]$.
From the common expression of variance, $$\operatorname{Var} [X] = \operatorname E[(X - \operatorname E[X])^2] = E[X^2] - \operatorname E [X]^2$$
if you accept that variance of any RV is intuitively $\ge 0$, or mathematically  $(X - \operatorname E[X])^2 \ge 0$ for all $X$ and then expected value of a non-negative RV is  non-negative (from a basic inequality on the  definition of EV as a sum/integral), you have the result.
Also $f(X) = X^2$ is a convex function, so Jensen's inequality also applies.
