# An Inequality with the Most Important Cevians

### Solution

As is common, we'll use $S,s,R,r$ for the area, the semiperimeter, the circumradius, the inradius of $\Delta ABC,$ respectively, and $a,b,c$ its side lengths.

From $ah_a=bh_b=ch_c=2S,$ since $abc=4RS,$

$\displaystyle h_ah_bh_c=\frac{8S^3}{abc}=\frac{8S^3}{4RS}=\frac{2S^2}{R}.$

From $m_a\ge\sqrt{s(s-a)},$ $m_b\ge\sqrt{s(s-b)},$ $m_c\ge\sqrt{s(s-c)},\,$

$\ell_a^2+\ell_b^2+\ell_c^2\le s(s-a)+s(s-b)+s(s-c)=s^2.$

From $\ell_a\ge h_a,$ $\ell_b\ge h_b,$ $\ell_c\ge h_c,$

\displaystyle \begin{align} \sum_{cycl}\ell_a\ell_b&\ge\sum_{cycl}h_ah_b=4S^2\sum_{cycl}\frac{1}{ab}\\ &=\frac{4S^2}{2Rr}=\frac{4s^2r^2}{2Rr}=\frac{2s^2r}{R}. \end{align}

Putting everything together:

$\displaystyle \frac{m_am_bm_c}{h_ah_bh_c}\ge\frac{R}{2r}\ge\frac{s^2}{2s^2r/R}\ge\frac{\ell_a^2+\ell_b^2+\ell_c^2}{\ell_a\ell_b+\ell_b\ell_c+\ell_c\ell_a}.$

### Acknowledgment

The problem is by Adil Abdullayev, the solution by Kevin Soto Palacios. I found the problem at the Peru Geometrico facebook group where it was posted by Miguel Ochoa Sanchez.