Let $Z=f(X_1,...X_n)$ where $X_1,...X_n$ are independant random variables taking their values on a measurable space $\chi$ and $f$ is a nonnegative function.
Let $Z_i=\inf_{x_i \in \chi} f (X_1, ... , X_{i-1}, x_i, X_{i+1}, ... , X_n)$ and $V=\sum_{i=1}^{n}\left ( Z-Z_i \right )^2$. I'm trying to prove that if there exists a random variable $W$ such that $V \leqslant WZ$, then $$\mathrm{Var}[\sqrt{Z}] \leqslant \mathrm{E}[W].$$
I managed to prove the weaker bound $\mathrm{Var}[\sqrt{Z}] \leqslant \sqrt{\mathrm{E}[W^2]}$. Here is how I obtained that result :
First, notice that the Hölder inequality $\mathrm{E}\left [ |XY| \right ] \leqslant \mathrm{E}\left [ |X|^{3/2} \right ]^{2/3}\mathrm{E}\left [ |Y|^3 \right ]^{1/3}$ applied to $X=Z^{1/3}$ and $Y=Z^{2/3}$ yields $$\mathrm{E\left [ \sqrt{Z} \right ]}^2 \geqslant \frac{\mathrm{E}\left [ Z \right ]^3}{\mathrm{E}\left [ Z^2 \right ]}.$$
This result gives in turn $$\mathrm{Var}[\sqrt{Z}] \leqslant \mathrm{Var}[Z]\frac{\mathrm{E}[Z]}{\mathrm{E}[Z^2]}.$$
Then, Efron-Stein's inequality and the fact that $V \leqslant WZ$ ensure that $\mathrm{Var}[Z] \leqslant \mathrm{E}[V] \leqslant \mathrm{E}[WZ]$, so that $$\mathrm{Var}[\sqrt{Z}] \leqslant \mathrm{E}[WZ]\frac{\mathrm{E}[Z]}{\mathrm{E}[Z^2]} \leqslant \sqrt{\mathrm{E}[W^2]\mathrm{E}[Z^2]}\frac{\mathrm{E}[Z]}{\mathrm{E}[Z^2]}=\sqrt{\mathrm{E}[W^2]}\frac{\mathrm{E}[Z]}{\sqrt{\mathrm{E}[Z^2]}}.$$ Since, $\mathrm{E}[Z] \leqslant \sqrt{\mathrm{E}[Z^2]}$, we get $\mathrm{Var}[\sqrt{Z}] \leqslant \sqrt{\mathrm{E}[W^2]}$.
My guess is that since $Z_i$ is defined in a specific way, its definition must be used in order to get a sharper bound, but I can't see how.