6. Given that $A, B, C$ are any three points, $BC=a, CA=b, AB=c$, then the minimum value of $y=\frac{c}{a+b}+\frac{b}{c}$ is
6. $\sqrt{2}-\frac{1}{2}$ Detailed explanation: From $a \leqslant b+c$, let $x=\frac{b}{c}$, then $y \geqslant \frac{c}{2 b+c}+\frac{b}{c}=\left(\frac{1}{2 x+1}+\frac{2 x+1}{2}\right)-\frac{1}{2} \geqslant \sqrt{2}-\frac{1}{2}$, when and only when $\frac{b}{c}=\frac{\sqrt{2}-1}{2}$ and $a=b+c$, $y$ has the minimum value $\sqrt{2}-\frac{1}{2}$.