<s>[INST] 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
 [/INST] Hint.
Making
$$
\cases{
u = x+y\\
v = x-y
}
$$
we have
$$
\sqrt{u^2-v^2}=2\frac uv
$$
so
$$
u^2 = \frac{v^4}{v^2-4}
$$
etc.
</s>