$V[\sqrt{X}]\le\frac{V[X]}{E[X]}$ for non-negative $X$ Let $X$ be a non-negative random variable with $0<E[X]<\infty$. Then the following inequality holds:$$V[\sqrt{X}]\le\frac{V[X]}{E[X]}$$
where $V$ stands for the variance.
This inequality is stated in a paper with no proof or reference.
Here are my thoughts: the inequality is homogeneous so we may suppose WLOG that $V[X]=1$ or that $V[\sqrt{X}]=1$. However I can't prove the inequality in either of these cases. Another line of thought is proving the equivalent inequality
$$2E[X]^2 \leq E[X^2] + E[X]E[\sqrt X]^2.$$
By Jensen's inequality we already have $E[X]^2\leq E[X^2]$ and so it would suffice to show $E[X]^2 \leq E[X]E[\sqrt X]^2$, i.e. $E[X]^{1/2}\leq E[\sqrt X]$. Unfortunately this does not hold in general (by Jensen's inequality and strict concavity of the square root it implies that $X$ is degenerate).
 A: Your problem essentially boils down to Exercise 3.19 of this book. That being said, let us suppose that $\mathrm{Var}(X) \leq a\mathbb E(X)$ for some $a \geq 0$. By Holder's inequality, we have $$\mathbb E(X) \leq \left(\mathbb E\sqrt{X}\right)^{\frac 23}\left(\mathbb EX^2\right)^{\frac 13} \leq \left(\mathbb E\sqrt{X}\right)^{\frac 23}\left((\mathbb EX)^2 + a\mathbb EX\right)^{\frac 13}.$$ Our goal is to show that $$\mathbb EX - \left(\mathbb E\sqrt{X}\right)^2 \leq a.$$ Now let us denote $v := \mathbb EX$ and $w = \mathbb E\sqrt{X}$, then the first inequality can be rewritten as $\frac{v^2}{v+a} \leq w^2$, so we end up with $$v - w^2 \leq v - \frac{v^2}{v+a} = \frac{va}{v+a} \leq a,$$ which is exactly what we wanted to prove.
A: Note: This answer is equivalent to this one (I have marked the linked answer as Community wiki now).
If you set $Y=\sqrt{X}$, then the inequality is equivalent to $$2\mathsf E(Y^2)^2\le \mathsf E(Y^2)\mathsf E(Y)^2+\mathsf E(Y^4).$$ Note that $Y\in L^4$ and $Y\ge 0$.
By AM-GM, $$\mathsf E(Y^2)\mathsf E(Y)^2+\mathsf E(Y^4)\ge 2\sqrt{\mathsf E(Y^2)\mathsf E(Y)^2\mathsf E(Y^4)},$$ so we only need to prove $$\mathsf E(Y^2)^4\le\mathsf E(Y^2)\mathsf E(Y)^2\mathsf E(Y^4),$$ i.e. $$\mathsf E(Y^2)^3\le \mathsf E(Y)^2\mathsf E(Y^4),$$ which is true by Hölder's inequality.
