An Inequality in Triangle, Mostly with the Medians

Problem

An Inequality in Triangle, Mostly with the Medians

Proof 1

By the AM_GM inequality, $\displaystyle \sqrt{(5m_a+3m_b)(3m_a+5m_b)}\le\frac{8(m_a+m_b)}{2},\,$ so that $(5m_a+3m_b)(3m_a+5m_b)\le 16(m_a+m_b)^2.\,$ Now, obviously, $\displaystyle m_a\lt\frac{b+c}{2},\,$ $\displaystyle m_b\lt\frac{c+a}{2},\,$ $\displaystyle m_c\lt\frac{a+b}{2},\,$ implying $16(m_a+m_b)^2\lt 4(a+b+2c)^2=4(2s+c)^2.\,$ It follows that

$(5m_a+3m_b)(3m_a+5m_b)\le 4(2s+c)^2.\,$

We similarly obtain

$(5m_b+3m_c)(3m_b+5m_c)\lt 4(2s+a)^2\,$ and
$(5m_c+3m_a)(3m_c+5m_a)\lt 4(2s+b)^2.\,$

The product of the three gives the required inequality.

Proof 2

First off, $2s+a=(b+a)+(c+a)\gt c+b\ge 2\sqrt{bc}\,$ so that

$\displaystyle \prod_{cycl}(2s+a)^2\gt \prod_{cycl}(2\sqrt{bc})^2=4^3(abc)^2.$

On the other hand,

$\displaystyle\begin{align} (5m_a+3m_b)(3m_a+5m_b) &= 15(m_a^2+m_b^2)+34m_am_b\\ &\le \frac{15}{4}(4c^2+a^2+b^2)+\frac{34}{2}(2c^2+ab), \end{align}$

because $\displaystyle m_a^2=\frac{2b^2+2c^2-a^2}{4},\,$ $\displaystyle m_b^2=\frac{2a^2+2c^2-b^2}{4}\,$ and $m_am_b\le\displaystyle\frac{2c^2+ab}{4}.\,$ We need to prove that

$\displaystyle 32c^2+\frac{15}{4}(a^2+b^2)+\frac{17}{2}ab\lt 4(2s+c)^2.$

This is equivalent to $\displaystyle 16c^2+\frac{ab}{2}\lt\frac{a^2+b^2}{4}+16c(a+b),\,$ which is true because $c\lt a+b\,$ and $2ab\le a^2+b^2.\,$ We only need to take the product of this and the two analogous inequalities to obtain the result.

Acknowledgment

Dan Sitaru has kindly posted at the CutTheKnotMath facebook page the above problem of his that was published in the Romanian Mathematical Magazine. Proof 1 is by Soumava Chakraborty; Proof 2 is by Diego Alvariz.

 

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