13. (Kazakhstan) Given a real number $A$, find the largest real number $M$ such that
$$\frac{1}{x}+\frac{1}{y}+\frac{A}{x+y} \geqslant \frac{M}{\sqrt{x y}}$$

holds for all $x, y > 0$.
Solution. Since both sides of the inequality are homogeneous, we may assume $xy=1, x+y=t$, then by the AM-GM inequality we have $t=x+y \geqslant 2 \sqrt{xy}=2$. The problem is transformed into, for a given real number $A$, finding the largest $M$ such that for any $t \geqslant 2$ we have $t+\frac{A}{t} \geqslant M$. Let $f(x)=x+\frac{A}{x}$, then $f^{\prime}(x)=1-\frac{A}{x^{2}}$.
(a) When $A \leqslant 4$, for $t \geqslant 2$, we always have $f^{\prime}(t)=1-\frac{A}{t^{2}} \geqslant 1-\frac{4}{t^{2}} \geqslant 0$. Therefore, in this case, the minimum value of $f(t)$ for $t \geqslant 2$ is $f(2)=2+\frac{A}{2}$, hence, the maximum value of $M$ is $2+\frac{A}{2}$.
(b) When $A>4$, it is easy to see that $f(t)$ is decreasing on $[2, \sqrt{A})$ and increasing on $[\sqrt{A},+\infty)$, so the minimum value of $f(t)$ for $t \geqslant 2$ is $f(\sqrt{A})^{2}=2 \sqrt{A}$, hence, the maximum value of $M$ is $2 \sqrt{A}$.