if $x-y =\sqrt{x}-\sqrt{y}$ with $x\neq y$ then $(1+\frac{1}{x})(1+\frac{1}{y})\geq 25$? let $x\neq y$ be positive real numbers such that :$x-y= \sqrt{x}-\sqrt{y}$ , I have tried to prove this inequality $(1+\frac{1}{x})(1+\frac{1}{y})\geq 25$ that i have created but i didn't got it.
Attempt I have showed that:$(\frac{1}{x}+\frac{1}{y})\geq \frac{2}{\sqrt{xy}}$ using this identity:  $(\sqrt{x}-\sqrt{y})^2\geq0$ , I also showed that :$\frac{1}{xy}\geq \frac{1}{16}$ , Now I have used both result I have got the following inequality :$(1+\frac{y}{x})(1+\frac{x}{y})\geq 25$ but not what i have claimed , any way ?

By AM-GM
$$\left(1+\frac{1}{x}\right)\left(1+\frac{1}{y}\right)=\left(1+\frac{4}{4x}\right)\left(1+\frac{4}{4y}\right)\geq$$
$$\geq\frac{5}{\sqrt[5]{(4x)^4}}\cdot\frac{5}{\sqrt[5]{(4y)^4}}=\frac{25}{\sqrt[5]{4^{8}\left(\sqrt{xy}\right)^8}}\geq\frac{5}{\sqrt[5]{4^8\left(\frac{\sqrt{x}+\sqrt{y}}{2}\right)^{16}}}=25.$$
