The inequality
$$\frac{3}{2} \le \frac{b+c}{b+c+2a} + \frac{a+c}{a+c+2b} + \frac{a+b}{a+b+2c}$$
is Nesbitt's inequality for the triple $(b+c, a+c, a+b)$.

Triangle's inequality yields $b+c > a \Leftrightarrow 3a+3b+3c > 4a+2b+2c \Leftrightarrow \frac{2}{3(a+b+c)} < \frac{1}{2a+b+c} \Leftrightarrow \frac{4a}{3(a+b+c)} < \frac{2a}{2a+b+c} \Leftrightarrow \frac{4a}{3(a+b+c)} < 1 - \frac{b+c}{2a+b+c}$.
In the same way $\frac{4b}{3(a+b+c)} < 1 - \frac{a+c}{2b+a+c}$ and $\frac{4c}{3(a+b+c)} < 1 - \frac{a+b}{2c+a+b}$. Adding the last three inequalities gives
$$\frac{4}{3} < 3 - \left( \frac{b+c}{b+c+2a} + \frac{a+c}{a+c+2b} + \frac{a+b}{a+b+2c} \right),$$
which is equivalent to what we had to prove.