Supremum of $\frac{(x+y)^{2}}{2^{xy}}$ How to find the supremum of 
$$
\frac{(x+y)^2}{2^{xy}}
$$
for $x\ge 1$ and $y\ge 1$
I have no idea how to start with it and when try to find it in WolframAlpha it only gives my numeric value.
 A: Hint On each line $x+y=c+1$ where $c\geq1$, the numerator is fixed and the denominator attains minimum when $x=1, y=c$.
A: Hints:  Let $f(x,y) = (x+y)^2/2^{xy}$.  There are three possibilities:  


*

*there exists an unbounded sequence of points $\{(x_n, y_n)\}$ (i.e. either $x_n$ or $y_n$ is going to positive infinity) such that $f(x_n, y_n)$ converges to the supremum of f,

*the supremum of f is achieved on the boundary where $x=1$ or $y=1$, or

*the supremum of f is achieved at a point $(x_1, y_1)$ where $x_1>1$, $y_1>1$, and the gradient at that point is the zero vector. 


When considering possibility 1., think about what happens when $x$ or $y$ is large.
When considering possibility 2, notice that $f(1,1) = 2$.  If you compute the gradient at (1,1), it indicates that the supremum does not occur at (1,1).  Alternatively, compute f(1.1,1.1) which is larger than 2.
The only way the maximum can occur on the line $x=1$ with $y>1$ is if the gradient is in the direction $(-1,0)$.  The only way the maximum can occur on the line $y=1$ with $x>1$ is if the gradient is in the direction $(0,-1)$.  
When considering possibility 3, the gradient must be the zero vector.  You can use that fact to restrict the possible locations of the maximum to a very simple set.      
That should be enough to get you started. 
Or, you could look at AdditIdent's answer which is a very big hint.
