[Math] Symmetry in inequalities.

inequality

I previously asked a question Is Symmetry a valid option in Inequalities.
After some thinking I concluded something:

Some theorem which I made up:[Call it my theorem :D]

Let $f$ be a diffrentiable,unbounded,multivariate function of n variables with real domain.If $f$ is symmetric*,
Then the minimum/maximum must occur at the point where all variables are equal,. For if $f$ is bounded the solutions may exist at the bounds.

Can need someone prove/disprove the statement, or prove a modified form of it with other minor restrictions.Best way would be some contracdictions, if it cannot be proved at all.

*By symmetric I mean all permutations of variables in the function give us the same expression.

I say so because, let the equation be:
$$f(x,y,z)=xy+yz+zx$$
It is symmetric because:
$$f(x,y,z)=f(x,z,y)=\cdots=f(z,y,x)$$

Now my thinking can be attributed to two further reasons:

  • If we get the minimum/maximum value of function at a particular set of values say $(\alpha,\beta,\gamma)$ then $\beta,\alpha,\gamma$ will also satisfy the equation. Letting all three values equal i.e. $\alpha=\beta=\gamma$ we might get a minimum/maximum.

  • $\displaystyle \frac{\partial f}{\partial x}=0,\frac{\partial f}{\partial y}=0,\frac{\partial f}{\partial z}=0$ all have same roots thus minimum/maximum might occur at point where all vaiables are equal.

Some more examples[All from MSE inequalitis section]:

  • $\displaystyle f=xyz-xy-yx-zx$, $x+y+z=1$ when $x=y=z=1/3$ minimum i.e. $-8/27$
  • $\displaystyle f=\frac{(x+y)(y+z)(z+x)}{(x+y+z)\sqrt[3]{x^2y^2z^2}}$,minimum when $x=y=z$ i.e. $8/3$
  • $\displaystyle f=\sum_{cyc} \frac{\sqrt{xy}}{\sqrt{xy+z}}\le\frac{3}{2}$, $x+y+z=1$ minimum when $x=y=z=1/3$ i.e. $3/2$
  • $\displaystyle f=a^3+b^3+c^3- a^2+b^2+c^2$, $ab+bc+ca\le 3abc$ minimum when $a=b=c$ i.e. $0$
  • $\displaystyle f=\sum{\frac{1}{(x+2y)^2}} -\frac{1}{xy+yz+zx}$, minimum when $x=y=z$ i.e. $0$
  • $\displaystyle f=(a+b+c)(1/a+1/b+1/c)$, minimum when $a=b=c$ i.e. $9$
  • $\displaystyle f=\sum_{cyc}\frac{(b+c-a)^4}{a(a+b-c)}-(ab+bc+ca)$, minimum when $a=b=c$ i.e. $0$

Best Answer

A modified version of this is definitely true:

Purkiss Principle:

Let $f$ and $g$ be symmetric functions with continuos second derivatives in the neighbourhood of a point $P=(r,r,r,...,r)$. On the set where $g$ equals $g(P)$, the function $f$ will have a local macimum or minimum at $P$ except in degenerate cases.

Link:

"Do symmetric problems have symmetric solutions?"
by William C.Waterhouse
Department of Mathematics, The Pennsylvania State University, University Park, PA 16802

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