Problem 925

13 Because
$$(b+c)^{2}=b^{2}+c^{2}+2 b c \geqslant 4 b c,$$

so $\frac{b+c}{b c} \geqslant \frac{4}{b+c}$, which means $\frac{1}{b}+\frac{1}{c} \geqslant \frac{4}{b+c}$. Since $a>0$, we have
$$\frac{a}{b}+\frac{a}{c} \geqslant \frac{4 a}{b+c},$$

Similarly, we get
$$\begin{array}{l} \frac{b}{c}+\frac{b}{a} \geqslant \frac{4 b}{c+a}, \\ \frac{c}{a}+\frac{c}{b} \geqslant \frac{4 c}{a+b}, \end{array}$$

Adding the above three inequalities, we obtain
$$\frac{b+c}{2 a}+\frac{c+a}{2 b}+\frac{a+b}{2 c} \geqslant \frac{2 a}{b+c}+\frac{2 b}{c+a}+\frac{2 c}{a+b} .$$

Three, 13. From $(b+c)^{2}=b^{2}+c^{2}+2 b c \geqslant 4 b c$
$$\Rightarrow \frac{b+c}{b c} \geqslant \frac{4}{b+c} \Rightarrow \frac{1}{b}+\frac{1}{c} \geqslant \frac{4}{b+c} \text {. }$$

Since $a>0$, we have $\frac{a}{b}+\frac{a}{c} \geqslant \frac{4 a}{b+c}$.
Similarly, $\frac{b}{c}+\frac{b}{a} \geqslant \frac{4 b}{c+a}, \frac{c}{a}+\frac{c}{b} \geqslant \frac{4 c}{a+b}$.
Adding the above three inequalities, we get
$$\frac{b+c}{2 a}+\frac{c+a}{2 b}+\frac{a+b}{2 c} \geqslant \frac{2 a}{b+c}+\frac{2 b}{c+a}+\frac{2 c}{a+b} \text {. }$$