Given positive $x,y$ such that $x > y$ and $\sqrt{x} \sqrt{y}(x-y) = x+y $, find minimum $(x+y)$ I am given positive numbers $x, y$ such that $x > y$ and $\sqrt{x} \sqrt{y}(x-y) = x+y $. I need to find the minimum value of $(x+y)$. Here is my try. Using AM-GM inequality for nonnegative numbers, I have
$$ 
\frac{(x+y)}{2} \geqslant \sqrt{x} \sqrt{y}  
$$
$$
 \sqrt{x} \sqrt{y}(x-y) \geqslant 2 \sqrt{x} \sqrt{y}   
 \\ \therefore (x-y) \geqslant 2  
$$
So, I have been able to arrive at this conclusion. But I am stuck here. Any help ?
Thanks

put $x=r^2{cos}^2a$  and $y=r^2{sin}^2a$ also let $a$ belong to $[0,\frac{\pi}{2}]$
thus we have to find max value of $r^2$
plugging the values in the given equation  and simplifying using basic trig formulas we have $r^4(cosa)(sina)(cos2a)=r^2$ or
$ r^2=\frac{4}{sin(4a)} \ge 4$
