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.
$$\frac{a}{b+c} + \frac{b}{a+c} + \frac{c}{a+b} = \frac{a^2}{ab+ac} + \frac{b^2}{ab+bc} + \frac{c^2}{ac+bc} \geq \frac{(a+b+c)^2}{2(ab+bc+ac)}$$
Recall Buniakowsky inequality: $$ (a^2+b^2+c^2)(x^2+y^2+z^2) \geq (ax+by+cz)^2$$ by expanding and regrouping the terms of: $$(ay-bx)^2 + (az-cx)^2 + (bz - cy)^2 \geq 0$$
So now: $$ \bigg (\frac{a^2}{x} + \frac{b^2}{y} + \frac{c^2}{z} \bigg )(x+y+z) \geq (a+b+c)^2, \forall x,y,z > 0$$ or $$ \frac{a^2}{x} + \frac{b^2}{y} + \frac{c^2}{z} \geq \frac{(a+b+c)^2}{x+y+z}, \forall x,y,z > 0$$
We have by AM-GM inequality twice: $\displaystyle \sum_{\text{cyclic}} \dfrac{a}{b+c}=\dfrac{1}{2}\displaystyle \sum_{\text{cyclic}} (a+b)\displaystyle \sum_{\text{cyclic}} \dfrac{1}{a+b}-3\geq \dfrac{1}{2}\cdot 3\cdot 3-3=\dfrac{3}{2}$
