Suppose that $f(x,y) \leq f(y,x)$ for all $(x,y)\in\textbf{R}^{2}$. Show that $f(x,y) = f(y,x)$ for all $(x,y)\in\textbf{R}^2$ The question is: Suppose that $f(x,y) \leq f(y,x)$ for all $(x,y)$ in $\textbf{R}^{2}$. Show that $f(x,y) = f(y,x)$ for all $(x,y)\in\textbf{R}^{2}$.
I tried searching this question, but all the things I found were about convex functions. We did not learn about that, or partial derivatives yet. The section this is in only encompasses limits of functions of two variables. I am wondering if the professor want us to use the squeeze theorem somehow to show it. Any explanation is greatly appreciated. 

If $f(x,y) \leq f(y,x)$ for all $(x,y)$ in $\textbf{R}^{2},$ 
then taking $(x,y)=(a,b)$ we have $f(a,b)\le f(b,a),$ 
and taking $(x,y)=(b,a)$ we have $f(b,a)\le f(a,b)$.  
So for all $a,b\in \textbf R, $ we have $f(a,b)\le f(b,a)\le f(a,b)$.  
Now if $f(a,b)<f(b,a)$, 
then we would have $f(a,b)<f(a,b),$ which is absurd.  
So it must be that $f(a,b)=f(b,a)$ for all $a,b\in \mathbf R.$
