Very easy.

OK.Here is my solution : Here, from apollonious theorem, we get- $ a^2 +b^2 = (\frac{c}{2})^2 + m_c^2......(1) $ $ a^2 +c^2 = (\frac{b}{2})^2 + m_b^2......(2) $ $ b^2 +c^2 = (\frac{a}{2})^2 + m_a^2......(3) $ By adding (1),(2),(3) & simply manipulation gives us desired result.

By Stewart's theorem $b^2\cdot\frac{a}{2}+c^2\cdot\frac{a}{2}=a\left(\left(\frac{a}{2}\right)^2+m_a^2\right)\iff 2b^2+2c^2=a^2+4m_a^2 $ Summing analogous equations $4(a^2+b^2+c^2)=(a^2+b^2+c^2)+4(m_a^2+m_b^2+m_c^2)^2$ qed

By median's formula for each get $4m_a^2=2(b^2+c^2)-a^2$ $4m_b^2=2(a^2+c^2)-b^2$ $4m_c^2=2(a^2+b^2)-c^2$ Sum up this equations and get desired result.

If someone would like to train above formula here is a task. Prove that $\frac{a^3+b^3+c^3}{a+b+c}\ge\frac{4}{9}(m_a^2+m_b^2+m_c^2)$

WolfusA wrote: If someone would like to train above formula here is a task. Prove that $\frac{a^3+b^3+c^3}{a+b+c}\ge\frac{4}{9}(m_a^2+m_b^2+m_c^2)$ Chebyshev

Prove that $\sum_{cyc}\frac{1}{a^2}\ge\frac{27}{4\sum_{cyc}m_a^2}$

WolfusA wrote: Prove that $\sum_{cyc}\frac{1}{a^2}\ge\frac{27}{4\sum_{cyc}m_a^2}$ The titu's lemma kills it

how to solve this problem WITHOUT stewart theorem????