<s>[INST] Prove: $\frac{a}{m_a}+\frac{b}{m_b}+\frac{c}{m_c}\ge 2\sqrt{3}$ Let $a,b,c$ be the lengths of the sides of triangle $ABC$ opposite $A,B,C$, respectively, and let $m_a,m_b,m_c$ be the lengths of the corresponding angle medians. Prove:
$$\frac{a}{m_a}+\frac{b}{m_b}+\frac{c}{m_c}\ge 2\sqrt{3}.$$
Source: I thought about it while solving the American Mathematical Monthly's problem 11945, which asked to prove $\frac{a}{w_a}+\frac{b}{w_b}+\frac{c}{w_c}\ge 2\sqrt{3}$, where $w_a,w_b,w_c$ are respective angle bisectors.
I think it can be solved by making up (inequality-) constrained optimization problem, but I am interested in solution within elementary geometry.
 [/INST] By AM-GM $$\sum_{cyc}\frac{a}{m_a}=\frac{2a}{\sqrt{2b^2+2c^2-a^2}}=\sum_{cyc}\frac{4\sqrt3a^2}{2\sqrt{3a^2(2b^2+2c^2-a^2)}}\geq$$
$$\geq\sum_{cyc}\frac{4\sqrt3a^2}{3a^2+2b^2+2c^2-a^2}=2\sqrt3\sum_{cyc}\frac{a^2}{a^2+b^2+c^2}=2\sqrt3.$$
Done!
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