Strictly convexity of $\rho \mapsto \int_{\mathbb{R}^N} |\nabla \sqrt{\rho}|^2 \,dx$ I want to know

Is the map $\rho \mapsto \int_{\mathbb{R}^N} |\nabla \sqrt{\rho}|^2 $ from $X \to \mathbb{R}$ strictly convex?
Here, $X = \{\rho\in W^{1,1}_+(\mathbb{R}^n)\mid \int \rho=1,\,\int |\nabla \rho|^2/\rho<\infty\}$ is arised from the minimization problem I mention in the last two paragraphs (Thanks to MaoWao's corrections).

I think this is true, that is, I want to show
\begin{align*}
\int_{\mathbb{R}^N} |\nabla \sqrt{\theta \rho_1 + (1-\theta) \rho_2}|^2 {\color{red} < } \int_{\mathbb{R}^N} \theta |\nabla \sqrt{\rho_1}|^2 + (1-\theta) |\nabla\sqrt{\rho_2}|^2
\end{align*} for any $0 < \theta <１$ and two $H^1$ functions $\rho_1 \neq \rho_2.$
Now I can only show this map is convex, that is,
\begin{align*}
\int_{\mathbb{R}^N} |\nabla \sqrt{\theta \rho_1 + (1-\theta) \rho_2}|^2 {\color{red} \leq } \int_{\mathbb{R}^N} \theta |\nabla \sqrt{\rho_1}|^2 + (1-\theta) |\nabla\sqrt{\rho_2}|^2
\end{align*}
and the equality holds if and only if
\begin{align}
|\nabla \log \sqrt{\rho_1}| = \frac{|\nabla \sqrt{\rho_1}|}{\sqrt{\rho_1}} {\color{red}\equiv} \frac{|\nabla \sqrt{\rho_2}|}{\sqrt{\rho_2}} = |\nabla \log \sqrt{\rho_2}|.
\end{align}
(One can show this easily by expanding the left-hand side and using the GM-AM inequality: $2|ab| \leq |a|^2 + |b|^2.$)(Some approximation procedure to avoid $\theta \rho_1(x) + (1-\theta) \rho_2(x) = 0$ in the denominator is also needed.)
I think this should imply $\rho_1 \equiv \rho_2.$ But I can't prove it.
The motivation for this problem is to show the uniqueness of the minimizer for the following minimization problem related to linear Schrodinger equation:
\begin{align*}
\inf\Big\{ \int_{\mathbb{R}^N} |\nabla u|^2 + V(x)u^2: u \in H^1(\mathbb{R}^N), \| u \|_{L^2} = 1 \Big\},
\end{align*}
where $0 \leq V$ and $V(x) \to \infty$ as $|x| \to \infty.$ The existence of minimizer for this problem is a consequence of compact embedding $H^1_V \subset\subset L^2,$ where $H^1_V$ norm is defined by $\| f \|_{H^1_V}^2:= \| f \|_{H^1}^2 + \int_{\mathbb{R}^N} V(x)f(x)^2.$
To deal with the uniqueness problem, it's natural to use $\sqrt{\rho} = u$ so that the nonconvex manifold ($L^2$ normalization condition) is transformed into a convex one. (Also thanks to the fact that this problem can be restricted to nonnegative functions $u \geq 0$ so that we can take square root.) So now the uniqueness problem is solved if I can show the desired strictly convexity.
 A: This functional is not strictly convex. Let $\rho_1,\rho_2$ have disjoint support. Then
\begin{align*}
\int|\nabla\sqrt {\theta\rho_1+(1-\theta)\rho_2)}|^2&=\int_{\operatorname{supp}\rho_1}|\nabla\sqrt{\theta\rho_1}|^2+\int_{\operatorname{supp}\rho_2}|\nabla\sqrt{(1-\theta)\rho_2}|^2\\
&=\theta\int|\nabla\sqrt{\rho_1}|^2+(1-\theta)\int|\nabla\sqrt{\rho_2}|^2.
\end{align*}
For many Schrödinger operators, unique continuation principles guarantee that eigenfunctions cannot be localized so that this failure of strict convexity does not matter for your minimization problem.
(If you don't restrict the functional to normalized $\rho$, then the failure of strict convexity is immediately clear because the functional is positively homogeneous).
However, in general your minimization problem does not have a unique solution. More precisely, let $K\subset \mathbb{R}^n$ a closed subset of measure zero. If $\mathbb{R}^n\setminus K$ is connected, then the ground state of $-\Delta+V$ is non-degenerate for all $V\in L^1_{\mathrm{loc}}(\mathbb{R}^n\setminus K)$, but if $\mathbb{R}^n\setminus K$ has $m$ connected for $m>1$, then the ground state of $-\Delta+V$ is $m$-times degenerate. This was proven by Faris and Simon in Degenerate and non-degenerate ground states for Schrödinger operators.
