I have a function $h(y,x_1,x_2,\ldots,x_n)$. It is known that the minimum value of $h$ for any $y$ is attained when $x_1 = x_n$ and $x_2 = x_3 = \cdots = x_{n-1}$. Now consider the following function \begin{equation} g(x_1,\ldots,x_n) = \int_{y\in\Theta}h(y,x_1,x_2,\ldots,x_n)f(y)dy \end{equation} where $f$ is some probability density function and $\Theta$ is appropriate space for $y$.

Numerically, I am getting that $g$ is also minimised when $x_1 = x_n$ and $x_2 = x_3 = \cdots = x_{n-1}$. However, analytically it is difficult to prove. Is there any result which ensures the optimal symmetry of solution even after taking the integration?

Edit 1: It is know that $h$ is a concave function of $(x_1,\ldots,x_n)$ and the vector $(x_1,\ldots,x_n)$ belongs to a convex set. Moreover, the density function is continuous (not discrete).

Edit 2: It is given that $(x_1,\ldots,x_n)\in\{(x_1,\ldots,x_n):x_i\geq 0,\sum_{i=1}^{n}x_i = 1\}$.