Equality

Prove that in any triangle $ABC$ it holds that

$16\sum m_a^2 m_b^2 = 9 \left ( s^2 + r^2 + 4Rr \right )^2 - 144 s^2 Rr$

Solution

We have successively:

$\begin{align*} 16 \sum m_a^2 m_b^2 &= 16 \sum \frac{2 \left ( b^2 + c^2 \right ) - a^2}{4} \cdot \frac{2 \left ( a^2 + c^2 \right ) - b^2}{4} \\ &= \sum \left ( 2b^2 + 2c^2 - a^2 \right ) \left ( 2a^2 + 2c^2 - b^2 \right ) \\ &= \sum \left ( 5a^2 b^2 + 2b^2 c^2 + 2a^2 c^2 - 2a^4 - 2b^4 + 4c^4 \right ) \\ &= 9 \left ( a^2 b^2 + b^2 c^2 + c^2 a^2 \right ) \\ &= 9 \left ( ab + bc + ca \right )^2 - 8abc \left ( a + b +c \right ) \\ &= 9 \left ( s^2 + r^2 + 4Rr \right )^2 - 144 Rrs^2 \end{align*}$