Question 52
If $a$, $b$ and $c$ are the length of the sides of a triangle, show that
$$
\frac{3}{2} \le \frac{b+c}{b+c+2a} + \frac{a+c}{a+c+2b} + \frac{a+b}{a+b+2c} \lt \frac{5}{3}.
$$
$$
\frac{3}{2} \le \frac{b+c}{b+c+2a} + \frac{a+c}{a+c+2b} + \frac{a+b}{a+b+2c} \lt \frac{5}{3}.
$$