11. (20 points) Find the largest real number $k$ such that for any triangle with sides $a$, $b$, and $c$, the following inequality holds:
$$
\frac{b c}{b+c-a}+\frac{a c}{a+c-b}+\frac{a b}{a+b-c} \geqslant k(a+b+c) .
$$
11. For an equilateral triangle, i.e., $a=b=c$, we have
$$
3 a \geqslant k \cdot 3 a \Rightarrow k \leqslant 1 \text {. }
$$

For any triangle, let
$$
x=\frac{b+c-a}{2}, y=\frac{c+a-b}{2}, z=\frac{a+b-c}{2} \text {. }
$$
$$
\begin{array}{l}
\text { Then } \frac{b c}{b+c-a}+\frac{c a}{c+a-b}+\frac{a b}{a+b-c} \\
=\frac{(x+y)(x+z)}{2 x}+\frac{(y+z)(y+x)}{2 y}+\frac{(z+x)(z+y)}{2 z} \\
=\frac{1}{2}\left(3 x+3 y+3 z+\frac{y z}{x}+\frac{z x}{y}+\frac{x y}{z}\right) .
\end{array}
$$

Notice that,
$$
\begin{aligned}
\frac{y z}{x} & +\frac{z x}{y}+\frac{x y}{z} \\
& =\frac{1}{2}\left(\left(\frac{y z}{x}+\frac{z x}{y}\right)+\left(\frac{z x}{y}+\frac{x y}{z}\right)+\left(\frac{x y}{z}+\frac{y z}{x}\right)\right) \\
& \geqslant z+x+y .
\end{aligned}
$$

Thus, $\frac{b c}{b+c-a}+\frac{c a}{c+a-b}+\frac{a b}{a+b-c}$
$$
\geqslant 2 x+2 y+2 z=a+b+c,
$$

which means $k=1$ satisfies the condition.
In conclusion, the maximum value of $k$ is 1.