13. Let $a, b>0$, find the largest positive integer $c$ such that for any positive real number $x$, we have $c \leqslant \max \{a x+ \left.\frac{1}{a x}, b x+\frac{1}{b x}\right\}$. (2003 Irish Mathematical Olympiad Problem)
13. Let $f(t)=t+\frac{1}{t}, t>0, g(x)=\max \{f(a x), f(b x)\}$. The problem is to find the minimum value of $g(x)$ for $x>0$.

If $a=b$, then $g(x)=a x+\frac{1}{a x} \geqslant 2$, with equality when $x=\frac{1}{a}$, in which case the desired $c=2$.
If $a \neq b$, without loss of generality, assume $0<f(t)$.

Therefore, when $x \geqslant \frac{1}{\sqrt{a b}}$, $f(b x) \geqslant f\left(\sqrt{\frac{b}{a}}\right)$; when $0<x \leqslant \frac{1}{\sqrt{a b}}$, we also have $f(b x) \geqslant f\left(\sqrt{\frac{a}{b}}\right)=f\left(\sqrt{\frac{b}{a}}\right)$.

Thus, $g(x) \geqslant f\left(\sqrt{\frac{b}{a}}\right)=\sqrt{\frac{b}{a}}+\sqrt{\frac{a}{b}}$. Equality is achieved when $x=\frac{1}{\sqrt{a b}}$.
In summary, the maximum $c=\sqrt{\frac{b}{a}}+\sqrt{\frac{a}{b}}$.