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?

Let $M,A,B,C$ be the geometric images of the complex numbers $z,z_{1},z_{2},z_{3}$,
let $f(x)=1$, then use Lagrange interpolation formula applied to the polynomial $f(x)$,
we have
$$\dfrac{(z-z_{1})(z-z_{2})}{(z_{3}-z_{1})(z_{3}-z_{2})}f(z_{3})+\dfrac{(z-z_{2})(z-z_{3})}{(z_{1}-z_{2})(z_{1}-z_{3})}f(z_{1})+\dfrac{(z-z_{3})(z-z_{1})}{(z_{2}-z_{3})(z_{2}-z_{1})}f(z_{2})=1$$
so
$$\dfrac{(z-z_{1})(z-z_{2})}{(z_{3}-z_{1})(z_{3}-z_{2})}+\dfrac{(z-z_{2})(z-z_{3})}{(z_{1}-z_{2})(z_{1}-z_{3})}+\dfrac{(z-z_{3})(z-z_{1})}{(z_{2}-z_{3})(z_{2}-z_{1})}=1$$
$$\Longrightarrow  \left|\dfrac{(z-z_{1})(z-z_{2})}{(z_{3}-z_{1})(z_{3}-z_{2})}\right|+\left|\dfrac{(z-z_{2})(z-z_{3})}{(z_{1}-z_{2})(z_{1}-z_{3})}\right|+\left|\dfrac{(z-z_{3})(z-z_{1})}{(z_{2}-z_{3})(z_{2}-z_{1})}\right|\ge 1$$
so
$$\dfrac{MA}{BC}\cdot\dfrac{MB}{CA}+\dfrac{MB}{CA}\cdot\dfrac{MC}{AB}+\dfrac{MC}{AB}\cdot\dfrac{MA}{BC}\ge 1$$
use this
$$(a+b+c)^2\ge 3(ab+bc+ac)$$
then we have
$$\dfrac{MA}{BC}+\dfrac{MB}{AC}+\dfrac{MC}{AB}\ge\sqrt{3}$$
