# 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.

Copyright © 1996-2018 Alexander Bogomolny

66887621