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. 

The thing is if something is true for all $(x,y)$ then it is true if we simply swap the positions of $x$ and $y$ and it will be true for $(y,x)$ as well.
If it helps you can put a second pair of variables in.  If it's true for any $x=x', y=y'$ then $f(x',y') \le f(y',x')$.  But then it is also true for $x=y'$ and $y=x'$ so $f(y',x') \le f(y',x')$.
But you shouldn't have to do that.
If $f(x,y) \le f(y,x)$ for all $x,y$ then $f(y,x) \le f(x,y)$ as well.  
So $f(x,y) \le f(y,x) $ and $f(y,x) \le f(x,y)$ for all $x,y$ so $f(y,x) =f(x,y)$ for all $x, y$.
