In a symmetric function, will equality always imply a maxima or minima? $w=f(x,y,z)=f(y,z,x)=f(z,x,y)$ for all positive reals $x,y,z$ that satisfy $x+y+z=3$.
$f$ is a rational function - ratio of two polynomial functions with real coefficients.
Is it possible that $f(1,1,1)$ is neither the maximum nor the minimum possible value of $w$?
I tried proving that $f(1,1,1)$ was either the maxima or minima using derivatives but was unable to do so. However, couldn't find any counter examples also.
P.s. I'm not that effluent at calculus, I only know basic differentiation and integration. I just thought of this using common sense, because often we end up using $a=b=c$ when solving a system containing an equality and an inequality.
 A: We often  "end up" with $x=y=z$ being the clue to the min or the max we are after if convexity is involved in some way. For a counterexample we therefore have to create a "nonconvex" situation. 
Consider the complex $w$-plane, $w=u+iv$ (this is not the $w$ in your question). The real-valued function
$$g(u,v):={\rm Re}(w^3)=u^3-3uv^2$$
is invariant under rotation by $120^\circ$ around the origin, and has a "monkey saddle" at $(0,0)$, see the following figure:

The maximum of $g$ on the equilateral triangle with vertices the third roots of $1$ is $g(1,0)=1$, and its minimum is $g\bigl(-{1\over2},0\bigr)=-{1\over8}$, whereas $g(0,0)=0$. Now we have to transport this $g$ somehow to the equilateral triangle  $\triangle\subset {\mathbb R}^3$ with vertices $(1,0,0)$, $(0,1,0)$, $(0,0,1)$. To this end consider the linear map
$$A:\quad{\mathbb R}^3\to{\mathbb R}^2,\qquad(x,y,z)\mapsto\left\{\eqalign{u&=x-{1\over2}y-{1\over2}z\cr v&={\sqrt{3}\over2}y-{\sqrt{3}\over2}z\cr}\right.\quad,$$
which maps the triangle $\triangle$ onto the triangle with vertices $(1,0)$, $\bigl(-{1\over2},\pm{\sqrt{3}\over2}\bigr)$. The function
$f(x,y,z):=g\bigl(u(x,y,z),v(x,y,z)\bigr)$ now has the desired properties. In particular
$$f(x,y,z)=x^3+y^3+z^3-{3\over2}(x^2y+x y^2+y^2 z+yz^2+z^2 x+x z^2)+6xyz$$ is symmetric, $f(1,0,0)=1$, $f\bigl(0,{1\over2},{1\over2}\bigr)=-{1\over8}$, $f\bigl({1\over3},{1\over3},{1\over3}\bigr)=0$, and $f\restriction\triangle$ has a monkey saddle at $\bigl({1\over3},{1\over3},{1\over3}\bigr)$.
A: The answer to your first question: $f:(x,y,z) \mapsto xyz$ is symmetric polynomial which has no local maxima/minima. It's not clear what $a,b,c$ have to do with it here.
