In a triangle ABC, $\cos \alpha =\frac{a}{b+c},\cos \beta =\frac{b}{c+a},\cos \gamma =\frac{c}{a+b}$ and  $k={\tan }^2\frac{\alpha }{2}\cdot {\tan }^2\frac{\beta }{2}\cdot {\tan }^2\frac{\gamma }{2}$. Then,

${\tan }^2\frac{\alpha }{2}=\frac{1-\cos \alpha }{1+\cos \alpha }=\frac{s-a}{s}$

Similarly, ${\tan }^2\frac{\beta }{2}=\frac{s-b}{s},{\tan }^2\frac{\gamma }{2}=\frac{s-c}{s}$

${\tan }^2\frac{\alpha }{2}+{\tan }^2\frac{\beta }{2}+{\tan }^2\frac{\gamma }{2}=1$; And by using

$\mathrm{AM}\ge \mathrm{GM}$.

$\begin{array}{l}{\tan }^2\alpha /2+{\tan }^2\beta /2+{\tan }^2\beta /2\ge \\ 3\cdot \sqrt[3]{{\tan }^2\frac{\alpha }{2}\cdot {\tan }^2\frac{\beta }{2}\cdot {\tan }^2\frac{\gamma }{2}}\\ \Rightarrow \sqrt[3]{\mathrm{k}}\le \frac{1}{3}\Rightarrow \mathrm{k}\le \frac{1}{27}\end{array}$