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.



Optimizing a multivariate symmetric (permutation-invariant) function
One can show that this function is *Schur-concave* using the *Schur-Ostrowski criterion*, which then implies the maximum is attained at the diagonal.


See also
<https://math.stackexchange.com/questions/439649/are-elementary-symmetric-polynomials-concave-on-probability-distributions>


I will omit the calculations here.


