Let $U, W \geq 0$, $U \leq W$ $\mathbb{P}( U < W) > 0$ and $\varepsilon > 0$ fix.
$\mathbf{Question:}$ Do we have $$ \mathbb{E}[(\varepsilon + U)^{1/2}]^2 - \mathbb{E}[W^{1/2}]^2 \leq \varepsilon $$ ?
$\mathbf{Ansatz:}$ By the third binomial formula, the left-hand side becomes $$ \mathbb{E}\left[(\varepsilon + U)^{1/2} + W^{1/2}\right]\mathbb{E}\left[(\varepsilon + U)^{1/2} - W^{1/2}\right]. $$ Since $\mathbb{E}[X]\mathbb{E}[Y] = \mathbb{E}[XY] - \text{cov}(X, Y)$, this equals \begin{align*} &\mathbb{E}\left[\left((\varepsilon + U)^{1/2} + W^{1/2} \right) \left((\varepsilon + U)^{1/2} - W^{1/2} \right)\right] - \left( \mathbf{Var}((\varepsilon + U)^{1/2}) - \mathbf{Var}(W^{1/2}) \right) \\ = &\mathbb{E}[(\varepsilon + U) - W] + \left( \mathbf{Var}(W^{1/2}) - \mathbf{Var}((\varepsilon + U)^{1/2}) \right) \\ =& \varepsilon + \mathbb{E}[U - W] + \left( \mathbf{Var}(W^{1/2}) - \mathbf{Var}((\varepsilon + U)^{1/2}) \right) \end{align*} This is smaller or equal than $\varepsilon$ if and only if $$ \mathbf{Var}(W^{1/2}) - \mathbf{Var}((\varepsilon + U)^{1/2}) \leq \mathbb{E}[W - U], $$ which does not seem to hold in general without any further assumptions on $U, W$. My assumption is therefore that the statement is wrong. I have tried to find counterexamples with two-point distributions but couldn't find one so far. I would therefore appreciate if someone managed to come up with a counterexample (or proof).
Thanks.