$B C=a, C A=b, A B=c$. Then $y=\frac{c}{a+b}+\frac{b}{c}$ 的最小值 is

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$B C=a, C A=b, A B=c$. Then $y=\frac{c}{a+b}+\frac{b}{c}$ the minimum value is
5. $\sqrt{2}-\frac{1}{2}$.

Given $b+c \geqslant a$.
$$
\begin{aligned}
\therefore y & =\frac{c}{a+b}+\frac{b}{c}=\frac{c}{a+b}+\frac{b+c}{c}-1 \\
& \geqslant \frac{c}{a+b}+\frac{a+b+c}{2 c}-1 \\
& =\frac{c}{a+b}+\frac{a+b}{2 c}-\frac{1}{2} \\
& \geqslant 2 \sqrt{\frac{c}{a+b} \cdot \frac{a+b}{2 c}}-\frac{1}{2}=\sqrt{2}-\frac{1}{2} .
\end{aligned}
$$

Equality holds if and only if $b+c=a$, and $\frac{a+b}{c}=\sqrt{2}$, i.e., $a=$ $\frac{\sqrt{2}+1}{2} c, b=\frac{\sqrt{2}-1}{2} c$. Therefore, $y_{\min }=\sqrt{2}-\frac{1}{2}$.