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 $$\dfrac{MA}{BC}+\dfrac{MB}{AC}+\dfrac{MC}{AB}\ge\sqrt{3}$$

• you're welcome= – math110 Jul 5 '15 at 4:33

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.

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 $\vec {GA}+\vec{GB}+\vec{GC} = \vec 0.$

• Nice proof with nice ideas! – Andreas Oct 6 '16 at 17:36