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? 
 A: There is no nice way. This is, however, the theory of symmetrization which attempts to show that for some extremal problems the solutions have certain symmetry. It goes back at least to Steiner symmetrization of 19th century. Much is known, but many open problems remain, and there is no universal theorem with which one could hammer problems without  looking at them. 
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.
A: Disclaimer: Certainly my sole purpose of this pseudo-answer is to send the OP to (a) right place. Apparently one nice tool to approach these type of problems, is so-called Purkiss principle.
Please see here.(answer provided by Henry Cohn) 
