<s>[INST] Show that $\frac{(m_a)^4+(m_b)^4+(m_c)^4}{a^4+b^4+c^4} = \frac{9}{16}$ Let $ m_a, m_b $ and $ m_c $ be the medians relative to the $ a, b, c $ sides of a triangle, show that:
 $$\frac{(m_a)^4+(m_b)^4+(m_c)^4}{a^4+b^4+c^4} = \frac{9}{16}$$
What i tried:
ust use Stewart’s theorem. We have $m=n=a/2,d=m_a$ so
$$\frac{a^3}{4}+m_a^2a=\frac{ab^2}{2}+\frac{ac^2}{2}\implies m_a^2=\frac{2b^2+2c^2-a^2}{4},$$and likewise for the other two medians. 
How to proceed?
 [/INST] $$\begin{align}
m_a^4+m_b^4+m_c^4&=\frac1{16}\sum_\text{cyc.}(2b^2+2c^2-a^2)^2\\
&=\frac1{16}\sum_\text{cyc.}4b^4+4c^4+a^4+8b^2c^2-4a^2b^ 2-4c^2a^2\\
&=\frac1{16}(9a^4+9b^4+9c^4).
\end{align}$$
</s>