Let $\ell$ and $d$ be two integers such that $\ell \le d$. I would like to find the global maxima of the following symmetric function $f\colon (0,1]^n \to \mathbb{R}$,

$$f(x_1, \ldots, x_n) := \sum_{\substack{S \subseteq [n] \\ |S| = \ell}} \sum_{\substack{z \in [d]^S \\ \sum_i z_i = d}} \prod_{i \in S} x_i \left(\ln\Bigl(\frac{e}{x_i^{1/z_i}}\Bigr)\right)^{z_i/2},$$

subject to $x_1 + \dots + x_n = 1$.

Note that this function has the same value under any permutation of the coordinates of $x$.

For example, when $n = 3, d = 2, \ell = 2$, we have $$f(x,y,z) = xy \sqrt{\ln(e/x) \ln(e/y)} + yz \sqrt{\ln(e/y) \ln(e/z)} + xz \sqrt{\ln(e/x) \ln(e/z)} .$$

When $n=3, d=3, \ell=2$, we have \begin{multline*} f(x,y,z) = xy \ln(e/\sqrt{x}) \sqrt{\ln(e/y)} + xy \sqrt{\ln(e/x)} \ln(e/\sqrt{y}) + yz \ln(e/\sqrt{y}) \sqrt{\ln(e/z)} + \\yz \sqrt{\ln(e/y)} \ln(e/\sqrt{z}) + xz \ln(e/\sqrt{x}) \sqrt{\ln(e/z)} + xz \sqrt{\ln(e/x)} \ln(e/\sqrt{z}) . \end{multline*}

The Purkiss Principle says that the point $(1/n, \ldots, 1/n)$ is a local maxima. I believe this is also the global maximum, but cannot think of a way to prove this.

One possible way is to show that this function is strictly concave, but it's not clear at all if this is true and I don't know how to show its Hessian is strictly negative-definite.

Is there an easy way to see this? or is my guess simply not true?

**Edit:** In the original question, the factor $\ln\Bigl(\frac{e}{x_i^{1/z_i}}\Bigr)$ was $\ln\Bigl(\frac{e}{x_i}\Bigr)$, which was clearly false as pointed out by @fedja.

notin the middle, which leaves little hope for the general case without additional assumptions. $\endgroup$ – fedja Dec 10 '18 at 20:58