Show that $\frac{(m_a)^4+(m_b)^4+(m_c)^4}{a^4+b^4+c^4} = \frac{9}{16}$

Question

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?

Answer 1

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,

$$\frac{m_a^4+m_b^4+m_c^4}{a^4+b^4+c^4} = \frac9{16}\frac{(a^2+b^2+c^2)^2-2(a^2b^2+b^2c^2+c^2a^2)}{a^4+b^4+c^4} =\frac{9}{16}$$

Answer 2

Use that $$m_a=\frac{1}{2}\sqrt{2(b^2+c^2)-a^2}$$ etc.

Answer 3

$$\begin{align} m_a^4+m_b^4+m_c^4&=\frac1{16}\sum_\text{cyc.}(2b^2+2c^2-a^2)^2\\ &=\frac1{16}\sum_\text{cyc.}4b^4+4c^4+a^4+8b^2c^2-4a^2b^ 2-4c^2a^2\\ &=\frac1{16}(9a^4+9b^4+9c^4). \end{align}$$