The inequality $\frac{MA}{BC}+\frac{MB}{CA}+\frac{MC}{AB}\geq \sqrt{3}$ Given a triangle $ABC$, and $M$ is an interior point. Prove that:
$\dfrac{MA}{BC}+\dfrac{MB}{CA}+\dfrac{MC}{AB}\geq \sqrt{3}$.
When does equality hold?

Another solution. Let $G$ be the centroid of the triangle $ABC$. It's well-known that $|GA|^2 = \dfrac{2(b^2+c^2)-a^2}{9}$, where $AB =c, BC =a, CA = b$. Thus, 
$$\sum_{a,b,c}\frac{MA}{a} = \sum_{a,b,c}\frac{\sqrt 3|MA||GA|}{\sqrt 3a|GA|}\geq\sum_{a,b,c}\dfrac{2\sqrt 3|MA||GA|}{a^2+3|GA|^2} = \sum_{a,b,c}\frac{3\sqrt 3|MA||GA|}{a^2+b^2+c^2} \\ \geq
3\sqrt 3\sum_{a,b,c}\frac{\vec {MA}\cdot\vec {GA}}{a^2+b^2+c^2} = 3\sqrt 3\sum_{a,b,c}\frac{|GA|^2+|GB|^2+|GC|^2}{a^2+b^2+c^2} = \sqrt 3$$, where in the last chain we have used the fact that $\vec {GA}+\vec{GB}+\vec{GC} = \vec 0.$
