# An Inequality with Sides and Medians

### Solution 1

Let, in the complex plane, $A=u,\,$ $B=v,\,$ $C=w.\,$ We have $a=|v-w|,\,$ $b=|w-u|,\,$ $c=|u-v|,\,$

\begin{align} 2m_a&=|v+w-2u|=|(w-u)-(u-v)|,\\ 2m_b&=|(u-v)-(v-w)|,\\ 2m_c&=|(v-w)-(w-u)|.\\ \end{align}

Introduce $w-u=x\,$ and $u-v=y.\,$ Then $v-w=-(x+y).\,$ Thus $a=|x+y|,\,$ $2m_a=|x-y|,\,$ $b=|x|,\,$ $2m_c=|2x+y|,\,$ $c=|y|,\,$ $2m_b=|x+2y|.$ Hence, we need to prove

$2|x+y||x-y|\le |x||2x+y|+|y||x+2y|,$

i.e., $2|x^2-y^2|\le |2x^2+xy|+|xy+2y^2|.\,$ Now, by the triangle inequality for modules,

$|2x^2+xy|+|xy+2y^2|\ge |(2x^2+xy)-(xy+2y^2)|=|2x^2-2y^2|,$

as required.

### Solution 2

Let $A'\,$ be the reflection of $A\,$ in the midpoint of $BC.$ Let $G\,$ be the centroid of $\Delta ABC.$

Then by Ptolemy's inequality in the quadrilateral $BA'CG,\,$ $BC\cdot A'G\le A'B\cdot CG+A'C\cdot BG,\;$ or more explicitly,

$\displaystyle a\cdot\frac{4}{3}m_a\le b\cdot\frac{2}{3}m_c+c\cdot\frac{2}{3}m_b,$

which exactly amounts to

$\displaystyle 2am_a\le bm_c+cm_b.$

### Solution 3

Let $B'\,$ be the midpoint of $AC,\,$ $C'\,$ that of $AB.\,$ Let $G\,$ be the centroid of $\Delta ABC.$

Apply Ptolemy's inequality in the quadrilateral $AB'GC':\,$ $GA\cdot B'C'\le AB'\cdot GC'+AC'\cdot GB',\,$ which translates into

$\displaystyle\frac{2m_a}{3}\cdot\frac{a}{2}\le\frac{b}{2}\cdot\frac{m_c}{3}+\frac{c}{2}\cdot\frac{m_b}{3},$

i.e., $2am_a\le bm_c+cm_b.$

### Acknowledgment

I am grateful to Leo Giugiuc who brought the above problem to my attention. The problem (by Dorin Andrica) is #J294 in the awesomemath.org set of problems for 2016. Solution 1 is by Leo Giugiuc; Solution 2 is by Gheorghe Duca; Solution 3 is by Marian Dinca.

[an error occurred while processing this directive] Copyright © 1996-2018 Alexander Bogomolny
[an error occurred while processing this directive]
[an error occurred while processing this directive]