How prove this inequality $a^2m_{b}m_{c}+b^2m_{c}m_{a}+c^2m_{a}m_{b}\ge\frac{1}{4}(a^2+b^2+c^2)^2$ 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
$$a^2m_{b}m_{c}+b^2m_{c}m_{a}+c^2m_{a}m_{b}\ge\dfrac{1}{4}(a^2+b^2+c^2)^2$$
$m_a^2 = \dfrac{2b^2+2c^2-a^2}{4}$
but I can't seem to arrive at the desired result.

We need to prove that 
$$\sum\limits_{cyc}m_a^2bc\geq\frac{4}{9}\left(\sum\limits_{cyc}m_a^2\right)^2$$ or 
$$\sum\limits_{cyc}(2b^2+2c^2-a^2)bc\geq(a^2+b^2+c^2)^2$$ or
$$\sum\limits_{cyc}(-a^4+2a^3b+2a^3c-2a^2b^2-a^2bc)\geq0$$ or
$$\sum\limits_{cyc}(-2a^4+4a^3b+4a^3c-4a^2b^2-2a^2bc)\geq0$$ or
$$\sum\limits_{cyc}(-2a^4+a^3b+a^3c+2a^3b+2a^3c-4a^2b^2+a^3b+a^3c-2a^2bc)\geq0$$ or
$$\sum\limits_{cyc}(a-b)^2(-a^2-b^2-ab+2ab+ac+bc)\geq0$$ or
$$\sum\limits_{cyc}(a-b)^2(c^2-a^2-b^2+2ab-c^2+ac+bc-ab)\geq0$$ or
$$\sum\limits_{cyc}(a-b)^2(c+a-b)(c+b-a)+\sum\limits_{cyc}(a-b)^2(c-a)(b-c)\geq0$$ or
$$\sum\limits_{cyc}(a-b)^2(c+a-b)(c+b-a)+(a-b)(b-c)(c-a)\sum\limits_{cyc}(a-b)\geq0$$ or
$$\sum\limits_{cyc}(a-b)^2(c+a-b)(c+b-a)\geq0$$
Done!
