Let $f:\mathbb R^n\longrightarrow [0,+\infty)$ be a $C^2$ function with compact support. Prove that $\sqrt f$ is $L$-Lipschitz, with $L^2\leq \frac{1}{2}\lambda(f)$, where $$\lambda(f)=\max_{x\in \mathbb R^n}\{\max_{\lambda_i\in \mathrm{Sp}(D^2f_x)} \{\lambda_i\}\}$$ (the greatest of all possible eigenvalues). Prove also that this estimation is optimal.
Hint: consider functions $f_\epsilon(x)=\sqrt{f(x)+\epsilon}$.
I tried Taylor expansion with first-order Lagrange remainder, which is of the form $\frac{1}{2}(x-y)^TD^2f_\xi (x-y)$, but the first-order term doesn't seem to vanish for any reason. Following the hint, I tried to differentiate $f_\epsilon$, which can be done since $f+\epsilon>0$, but the first order term, which should give me the requested constant, has nothing to do with $D^2 f$: $$f_\epsilon(x)=f_\epsilon(y)+\frac{Df_y}{2\sqrt{f(y)+\epsilon}}\cdot (x-y)+O(|x-y|^2)$$ (and the Lagrange remainder doesn't either seem to help).
Also general suggestions of helpful tools are welcome.
This is really a one-dimensional problem: to verify the Lipschitz condition for a pair of points, it suffices to consider the restriction of $f$ (denoted $g$) to the line through these points. The assumptions on the eigenvalues then becomes simple: $g''\le \lambda(f)$.
Let $t_0$ be any point of $\mathbb{R}$. The tangent line to the graph of $g$ has equation $\ell(t) = g(t_0)+g'(t_0)(t-t_0)$. Consider also the parabola $$p(t) = g(t_0)+g'(t_0)(t-t_0)+\frac12\lambda(f) (t-t_0)^2$$ Since $p(t_0)=g(t_0)$, $p'(t_0)=g'(t_0)$ and $p''\ge g''$ everywhere, it follows that $p\ge g$ everywhere on $\mathbb{R}$.
The assumption $g\ge 0$ then implies $p\ge 0$ on $\mathbb{R}$. In terms of the discriminant of the quadratic polynomial $p$, this yields $g'(t_0)^2 \le 2\lambda(f)g(t_0)$. Since $t_0$ was arbitrary, we conclude that $$|g'|\le \sqrt{2\lambda(f)} {\sqrt{g}} \quad \text{on } \ \mathbb{R}\tag1$$
Formally, $|g'|/\sqrt{g} = 2|h'|$ where $h=\sqrt{g}$, so the desired Lipschitz bound follows... except for the issue with the points where $g=0$. This can be overcome in a few ways, such as $+\epsilon$ trick that you were suggested.