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?

Note,
$$m_a^4+m_b^4+m_c^4 = (m_a^2+m_b^2+m_c^2 )^2 - 2(m_a^2m_b^2+m_b^2m_c^2+m_c^2m_a^2) $$
Then, evaluate,
$$m_a^2+m_b^2+m_c^2 = \frac{2b^2+2c^2-a^2}{4} +\frac{2c^2+2a^2-b^2}{4} +\frac{2a^2+2b^2-c^2}{4} = \frac34 (a^2+b^2+c^2)$$
$$m_a^2m_b^2+m_b^2m_c^2+m_c^2m_a^2 = \frac9{16}(a^2b^2+b^2c^2+c^2a^2)$$
Thus,
$$\frac{m_a^4+m_b^4+m_c^4}{a^4+b^4+c^4} = \frac9{16}\frac{(a^2+b^2+c^2)^2-2(a^2b^2+b^2c^2+c^2a^2)}{a^4+b^4+c^4} =\frac{9}{16}$$
