Prove that for positive real numbers $a,b,c$ we have $\frac{a}{b+c}+ \frac{b}{a+c}+\frac{c}{a+b} \geq \frac{3}{2}.$ 
Prove that for positive real numbers $a,b,c$ we have $$\dfrac{a}{b+c}+ \dfrac{b}{a+c}+\dfrac{c}{a+b} \geq \dfrac{3}{2}.$$

Attempt
I tried using AM-GM and got $ \dfrac{a}{2\sqrt{bc}}+\dfrac{b}{2\sqrt{ac}}+\dfrac{c}{2\sqrt{ab}} \geq \dfrac{a}{b+c}+ \dfrac{b}{a+c}+\dfrac{c}{a+b}$ but that doesn't seem to help since that gives an upper not lower bound.

We have by AM-GM inequality twice: $\displaystyle \sum_{\text{cyclic}} \dfrac{a}{b+c}=\dfrac{1}{2}\displaystyle \sum_{\text{cyclic}} (a+b)\displaystyle \sum_{\text{cyclic}} \dfrac{1}{a+b}-3\geq \dfrac{1}{2}\cdot 3\cdot 3-3=\dfrac{3}{2}$
