2. Given $A, B, C$ are any three points on a plane, and $BC=a, CA=b, AB=c, y=$ $\frac{c}{a+b}+\frac{b}{c}$, the minimum value of $y$ is . $\qquad$
$\begin{array}{l}\text { 2. } \sqrt{2}-\frac{1}{2} . \\ b+c \geqslant a, y=\frac{c}{a+b}+\frac{b}{c}=\frac{c}{a+b}+\frac{b+c}{c}-1 \geqslant \frac{c}{a+b}+\frac{a+b+c}{2 c}-1=\frac{c}{a+b}+\frac{a+b}{2 c}- \\ \frac{1}{2} \geqslant 2 \sqrt{\frac{c}{a+b} \cdot \frac{a+b}{2 c}}-\frac{1}{2}=\sqrt{2}-\frac{1}{2} \text {, when and only when } a=\frac{\sqrt{2}+1}{2} c, b=\frac{\sqrt{2}-1}{2} c \text { the equality holds. }\end{array}$