# An Inequality in Triangle XI

### Solution

Let $G$ denote the centroid of $\Delta ABC$.

$\displaystyle AG=\frac{2}{3}m_a;\;BG=\frac{2}{3}m_b$

\displaystyle \begin{align} 1&\gt \cos (\widehat{GBA})=\frac{GB^2+AB^2-GA^2}{2GB\cdot AB}\\\\ &=\frac{\displaystyle (\frac{2}{3}m_b)^2+c^2-(\frac{2}{3}m_a)^2}{\displaystyle 2\cdot \frac{2}{3}m_b\cdot c}\\\\ &=\frac{\displaystyle 9c^2+4m_b^2-4m_a^2}{\displaystyle 12cm_b}\\\\ &=\frac{\displaystyle 9c^2+2a^2+2c^2-b^2-2b^2-2c^2+a^2}{\displaystyle 12cm_b}\\\\ &=\frac{\displaystyle 9c^2+3a^2-3b^2}{\displaystyle 12cm_b}=\frac{\displaystyle 3c^2+a^2-b^2}{\displaystyle 4cm_b}. \end{align}

From here, $3c^2+a^2-b^2\lt 4cm_b$ and, similarly, $3a^2+b^2-c^2\lt 4am_c$ and $3b^2+c^2-a^2\lt 4bm_a.$ By adding the three,

$3(a^2+b^2+c^2)\lt 4(am_c+bm_a+cm_b).$

### Acknowledgment

I am grateful to Dan Sitaru who kindly communicated to me this problem, along with a solution of his, all in a LaTeX file which I very much appreciate.