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?
 A: Geometric Proof:
In this proof, $M$ is not assumed to be inside $ABC$.  Let $P$ and $Q$ be the points such that $APBC$ and $MQBC$ are parallelograms.  Note that $APQM$ is also a parallelogram.  Hence, $AP=BC=MQ$, $BP=CA$, $BQ=MC$, and $PQ=MA$.
Consider the quadrilateral $APQB$.  We have by Ptolemy's Inequality that $$MC\cdot BC+MA\cdot AB=AP\cdot BQ+PQ\cdot AB\geq BP\cdot AQ =CA\cdot AQ\,.$$
Hence,
$$MB\cdot MC\cdot BC+MA \cdot MB\cdot AB\geq CA\cdot MB\cdot AQ \,,$$
and so
$$MB\cdot MC\cdot BC+MC\cdot MA\cdot CA+MA \cdot MB\cdot AB \geq CA\cdot (MC\cdot MA+MB\cdot AQ)\,.$$
Now, look at the quadrilateral $MAQB$.  By Ptolemy's Inequality,
$$MC\cdot MA+MB\cdot AQ=BQ\cdot MA+MB\cdot AQ\geq MQ\cdot AB=BC\cdot AB\,. $$
That is, 
$$MB\cdot MC\cdot BC+MC\cdot MA\cdot CA+MA \cdot MB\cdot AB \geq CA\cdot(BC\cdot AB)=BC\cdot CA\cdot AB\,,$$
whence
$$\frac{MB}{CA}\cdot\frac{MC}{AB}+\frac{MC}{AB}\cdot\frac{MA}{BC}+\frac{MA}{BC}\cdot\frac{MB}{CA}\geq 1\,.$$
The equality holds if and only if both $APQB$ and $MAQB$ are convex cyclic quadrilaterals, which means that $M\in\{A,B,C\}$ or that, if $ABC$ is an acute triangle, $M$ is the orthocenter of $ABC$.
The rest is as in math110's solution.  The inequality $\frac{MA}{BC}+\frac{MB}{CA}+\frac{MC}{AB}\geq\sqrt{3}$ becomes an equality if and only if $ABC$ is equilateral and $M$ is its centroid.
A: 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 $\vec {GA}+\vec{GB}+\vec{GC} = \vec 0.$
A: 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
$$\dfrac{MA}{BC}+\dfrac{MB}{AC}+\dfrac{MC}{AB}\ge\sqrt{3}$$
