# Minima of symmetric functions given a constraint

If $f(x,y,z,\ldots)$ is symmetric in all variables, (i.e $f$ remains the same after interchanging any two variables), and we want to find the extrema of $f$ given a symmetric constraint $g(x,y,z,\ldots)=0$,

$$\bf\text{When is it true that the extrema is achieved when }\ x=y=z=\ldots?$$

An example where this claim is true: $$g(x,y,z) = x+y+z - xyz,\ f(x,y,z) =(x-1)(y-1)(z-1)$$ An example where this claim is false (courtesy of @N.S): $$g(x,y,z) = x^2+y^2+z^2-1, \ f(x,y,z) = x^4+y^4+z^4$$

Is there a nice way to know when symmetry can be used to claim that the extrema is achieved when all variables are equal?

• Symmetry only tells you the feasible set is symmetric by permutation of the variables and so are the level sets of the objective function. But now imagine sets that are symmetric by permutation of the axes. There is no reason to expect that the optimum points are going to be on the diagonal (all variables equal). Optimum points could just be located at symmetric points that get permuted after permutation of the axes. There is no easy characterization for this. – OR. Jul 20 '13 at 19:42

Consider the following special case of your problem: maximize $\sum_{k=1}^n \psi(x_k)$ subject to $0\le x_k\le 1$ for all $k$, and $\sum_{k=1}^n x_k=1$. Given a nice smooth function $\psi:[0,1]\to \mathbb R$, how can one tell whether the maximum is attained at $x_1=\dots=x_n=1/n$? Surely the maximum is there, when $\psi$ is concave on $[0,1]$. And it's not there if $\psi$ is strictly convex in a neighborhood of $1/n$. But when $\psi$ is concave around $1/n$, yet convex somewhere else on the interval, the problem is nontrivial even if you have an explicit form of $\psi$. The solution will likely involve sweating through estimates, cases and subcases.