If $a, b$ and $c$ are the length of the sides of a triangle, show that
$$\frac32 \le \frac{b + c}{b + c + 2a}+
\frac{a + c}{a + c + 2b}+
\frac{a + b}{a + b + 2c}\le \frac53.$$
Use Radon's Inequality for both sides. Define the sum as $S$.
$$S=\sum_{cyc}{\left(\dfrac{b}{b+c+2a}+\dfrac{c}{b+c+2a}\right)}\geq \dfrac{2(a+b+c)^2}{a^2+b^2+c^2+3(ab+bc+ca)}\geq \dfrac{3}{2}$$For the other side, it sufficies to prove
$$3-S=\sum_{cyc}{\dfrac{2a}{b+c+2a}}\leq \dfrac{4}{3}$$But
$$\sum_{cyc}{\dfrac{2a}{b+c+2a}}\geq \dfrac{(a+b+c)^2}{a^2+b^2+c^2+ab+bc+ca}\geq \dfrac{4}{3}$$$$\Longleftrightarrow a(b+c-a)+b(c+a-b)+c(a+b-c)\geq 0$$as desired.