I have a square-integrable random variable $X > 0$. Function $f : \mathbb{R}^+ \rightarrow \mathbb{R}^+ $ is concave, monotonically increasing and $f(0) = 0$. I would like to prove that (and make sure that it is actually true): \begin{equation} \frac{\mathbf{E}[X^2]}{\mathbf{E}[X]^2} \geq \frac{ \mathbf{E}[f(X)^{2}]}{ \mathbf{E} [f(X)]^2} \end{equation} Which is equivalent to saying that the coefficient of variation ($c_V$, CV) decreases after applying $f$ to $X$. Given of course that $\mathbf{E}[f(X)^{2}]$ and $\mathbf{E} [f(X)]$ exist.
It seems to be true numerically and this Prove $\mathbf{E}[X^2] \geq \frac{ \mathbf{E}[X^{2\alpha}]}{ \mathbf{E} [X^{\alpha}]^2}$ seems to be a special case, but I can't come up with the analytical proof. It appears quite related to Jensen's inequality, but I can't use Hölder's inequality as in Prove $\mathbf{E}[X^2] \geq \frac{ \mathbf{E}[X^{2\alpha}]}{ \mathbf{E} [X^{\alpha}]^2}$. Also, maybe the condition $f(0) = 0$ might not be needed for this to be true. Thanks a lot of your help. Any leads would be highly appreaciated.