# An Inequality in Triangle with Differences of the Medians

### Solution 1

\displaystyle\begin{align} (m_a-m_b)(m_a+m_b) &= m_a^2-m_b^2\\ &=\frac{1}{2}(b^2+c^2)-\frac{1}{4}a^2-\frac{1}{2}(a^2+c^2)+\frac{1}{4}b^2\\ &=\frac{2b^2+2c^2-a^2-2a^2-2c^2+b^2}{4}\\ &=\frac{3(b^2-a^2)}{4}=\frac{3(b-a)(b+a)}{4}\\ &\frac{m_a-m_b}{b-a}=\frac{3(b+a)}{4(m_a+m_b)}>\frac{3(b+a)}{4(\frac{b+c}{2}+\frac{a+c}{2})}\\ &=\frac{3(b+a)}{2(a+b+2c)}=\frac{3(b+a)}{2(2s+c)}\geq \frac{3\cdot 2\sqrt{ab}}{2(2s+c)} \end{align}

It follows that

$\displaystyle\frac{2(m_a-m_b)}{b-a}>\frac{3\sqrt{ab}}{2s+c}.$

Similarly, $\displaystyle\frac{2(m_b-m_c)}{c-b}>\frac{3\sqrt{bc}}{2s+a}\,$ and $\displaystyle\frac{2(m_c-m_a)}{a-c}>\frac{3\sqrt{ca}}{2s+b}.\,$ Multiplying the three relationships yields

$\displaystyle\frac{8(m_a-m_b)(m_b-m_c)(m_c-m_a)}{(b-a)(c-b)(a-c)}>\frac{27abc}{(a+2s)(b+2s)(c+2s)}.$

### Solution 2

First we note $a\gt b\;\Rightarrow\,m_a\lt m_b.\,$ Indeed, from $m_a^2=\displaystyle\frac{b^2+c^2}{2}-\frac{a^2}{4}\,$ and $m_b^2=\displaystyle\frac{a^2+c^2}{2}-\frac{b^2}{4}\,$ we obtain $\displaystyle m_a^2-m_b^2=\frac{3}{4}(b^2-a^2).$

Now, using $\displaystyle m_am_b\le\frac{2c^2+ab}{4},$

\displaystyle\begin{align} (m_a-m_b)^2 &= m_a^2+m_b^2-2m_am_b\\ &\ge\frac{a^2+b+2+4c^2}{4}-\frac{2c^2+ab}{2}\\ &=\frac{(b-a)^2}{4}, \end{align}

such that $\displaystyle \frac{2(m_a-m_b)}{b-a}\ge 1.\,$ Similarly, $\displaystyle \frac{2(m_b-m_c)}{c-b}\ge 1\,$ and $\displaystyle \frac{2(m_c-m_a)}{a-c}\ge 1\,$ the product of which leads to

$\displaystyle \frac{8(m_a-m_b)(m_b-m_c)(m_c-m_a)}{(b-a)(c-b)(a-c)}\ge 1.$

Suffice it to prove that $1\gt \displaystyle\frac{27abc}{(a+2s)(b+2s)(c+2s)}.\,$ But, by the AM-GM inequality, $a+b+c\ge 3\sqrt[3]{abc}.\,$ Thus, we continue

\displaystyle\begin{align} \frac{27abc}{(a+2s)(b+2s)(c+2s)}&\lt \frac{27abc}{(2s)(2s)(2s)}\\ &=\frac{27abc}{(a+b+c)^3}\\ &\le\frac{27abc}{27abc}=1. \end{align}

This completes the proof.

### Acknowledgment

Dan Sitaru has kindly posted at the CutTheKnotMath facebook page the above problem of his that was published in the Romanian Mathematical Magazine. Dan messaged me his solution in a tex file and later added two more solutions. Solution 2 is by Soumava Chakraborty. Serban George Florin and independently Athina Kalampoka and Chris Kyriazis gave solutions very similar to that of Dan Sitaru.