<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] Given $\sqrt{xy}\left(x-y\right)=x+y$
let $yx=c$ , where $c>0$.
$$\sqrt{c}\left(x^{2}-c\right)=x^{2}+c$$
$$x^{2}-\frac{\left(c+1+2\sqrt{c}\right)}{\left(c-1\right)}c=0 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ -[1]$$
Let a function $$F(x,c)=x^{2}-\frac{\left(c+1+2\sqrt{c}\right)}{\left(c-1\right)}c$$ be defined. Then $$\frac{\partial F(x,c)}{\partial c}=0$$ at a constant $x$ gives us $c \approx 2.618 \implies x \approx 3.33 $ ( using $[1]$ ) . So, $$x+\frac{c}{x}\geqslant 4$$
$$min(x+y)=4$$
when $x=2+\sqrt{2} \text{ and } y=2-\sqrt2$ as pointed out.
</s>