Given positive $x,y$ such that $x > y$ and $\sqrt{x} \sqrt{y}(x-y) = x+y $, find minimum $(x+y)$

Question

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

Answer 1

By AM-GM $$(x+y)^2=xy(x-y)^2=\frac{1}{4}\cdot4xy(x-y)^2\leq\frac{1}{4}\left(\frac{4xy+(x-y)^2}{2}\right)^2=\frac{(x+y)^4}{16},$$ which gives $$x+y\geq4.$$ The equality occurs for $(x-y)\sqrt{xy}=x+y$ and $4xy=(x-y)^2,$ which gives $$(x,y)=(2+\sqrt2,2-\sqrt2),$$ which says that we got a minimal value.