# An Inequality in Triangle, with Sides and Medians

### Solution

We have a self-explanatory sequence of steps:

\displaystyle \begin{align} &GB+GC\gt BC\\ &GB+GC\gt a\Rightarrow \frac{GB+GC}{GA}\gt \frac{a}{GA}\\ &\Bigr(\frac{GB+GC}{GA}\Bigr)^4\gt \Bigr(\frac{a}{GA}\Bigr)^4\\ &\Bigr(\frac{GB}{GA}+\frac{GC}{GA}\Bigr)^4\gt \Bigr(\frac{a}{\frac{2}{3}m_a}\Bigr)^4\\ &\Biggl(\frac{\frac{2}{3}m_b}{\frac{2}{3}m_a}+\frac{\frac{2}{3}m_c}{\frac{2}{3}m_a}\Biggl)^4\gt \frac{81}{16}\Bigr(\frac{a}{m_a}\Bigr)^4\\ &\Bigr(\frac{m_b}{m_a}+\frac{m_c}{m_a}\Bigr)^4\gt \frac{81}{16}\Bigr(\frac{a}{m_a}\Bigr)^4\\ &16\Bigr(\frac{m_b}{m_a}+\frac{m_c}{m_a}\Bigr)^4\gt 81\Bigr(\frac{a}{m_a}\Bigr)^4\\ &16\sum \Bigr(\frac{m_a}{m_c}+\frac{m_b}{m_c}\Bigr)^4\gt 81\Biggl(\Bigr(\frac{a}{m_a}\Bigr)^4+\Bigr(\frac{b}{m_b}\Bigr)^4+\Bigr(\frac{c}{m_c}\Bigr)^4\Biggl) \end{align}

### Acknowledgment

The problem from the Romanian Mathematical Magazine has been kindly posted by Dan Sitaru at the CutTheKnotMath facebook page. Dan also has communicated his solution in a latex file.