Prove that $m_a(bc-a^2)+m_b(ac-b^2)+m_c(ab-c^2) \geq 0.$

Question

If $a,b,$ and $c$ are the side lengths of a triangle, $m_a, m_b,$ and $m_c$ are the lengths of the medians, prove that $$m_a(bc-a^2)+m_b(ac-b^2)+m_c(ab-c^2) \geq 0.$$

I was thinking of using power of a point and the median formula $m_a^2 = \dfrac{2b^2+2c^2-a^2}{4}$, but I can't seem to arrive at the desired result.

Answer 1

$\sum\limits_{cyc}m_a(bc-a^2)=\frac{1}{2}\sum\limits_{cyc}m_a((c-a)(a+b)-(a-b)(a+c))=$

$=\frac{1}{2}\sum\limits_{cyc}(a-b)((b+c)m_b-(a+c)m_a)=\frac{1}{8}\sum\limits_{cyc}\frac{(a-b)((b+c)^2(2a^2+2c^2-b^2)-(a+c)^2(2b^2+2c^2-a^2))}{(b+c)m_b+(a+c)m_a}=$

$=\frac{1}{8}\sum\limits_{cyc}\frac{(a-b)^2(a^3+a^2b+ab^2+b^3+2c(a^2+3ab+b^2)+c^2(a+b)-4c^3)}{(b+c)m_b+(a+c)m_a}\geq$

$\geq\frac{1}{8}\sum\limits_{cyc}\frac{(a-b)^2(a^3+abc+b^3+2c^3+2abc+c^3-4c^3)}{(b+c)m_b+(a+c)m_a}=\frac{1}{8}\sum\limits_{cyc}\frac{(a-b)^2(a^3+b^3-c^3+3abc)}{(b+c)m_b+(a+c)m_a}=$

$=\frac{1}{8}\sum\limits_{cyc}\frac{(a-b)^2(a+b-c)(a^2+b^2+c^2-ab+ac+bc)}{(b+c)m_b+(a+c)m_a}\geq0$