13. $[8]$ Let $a, b$, and $c$ be the side lengths of a triangle, and assume that $a \leq b$ and $a \leq c$. Let $x=\frac{b+c-a}{2}$. If $r$ and $R$ denote the inradius and circumradius, respectively, find the minimum value of $\frac{a x}{r R}$.
Answer: 3
It is well-known that both $\frac{a b c}{4 R}$ and $\frac{r(a+b+c)}{2}$ are equal to the area of triangle $A B C$. Thus $\frac{a b c}{4 R}=\frac{r(a+b+c)}{2}$, and
$$
R r=\frac{a b c}{2(a+b+c)} .
$$

Since $a \leq b$ and $a \leq c$, we have $\frac{a^{2}}{b c} \leq 1$. We thus obtain that
$$
\begin{aligned}
\frac{a x}{r R} & =\frac{a(b+c-a) / 2}{\frac{a b c}{2(a+b+c)}} \\
& =\frac{(a+b+c)(b+c-a)}{b c} \\
& =\frac{(b+c)^{2}-a^{2}}{b c} \\
& =\frac{(b+c)^{2}}{b c}-\frac{a^{2}}{b c} \\
& =\frac{b}{c}+\frac{c}{b}+2-\frac{a^{2}}{b c} \\
& \geq \frac{b}{c}+\frac{c}{b}+2-1 \\
& \geq 2+2-1 \\
& =3
\end{aligned}
$$

Equality is achieved when $a=b=c$.