An Inequality in Triangle with Differences of the Medians

Problem

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.

 

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