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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.

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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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  • $\begingroup$ nice solution. I started this way: due to normalization $a^2+b^2+c^2=1$. The inequality is $$\sum_{cyc} \frac{2a}{\sqrt{(a^2+b^2+c^2)(2b^2+2c^2-a^2)}}\stackrel{AM-GM}\ge\sum_{cyc} \frac{4a}{3(b^2+c^2)}\ge2\sqrt{3}.$$ $\endgroup$ – farruhota May 12 '17 at 11:12
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We need to prove that $$\sum_{cyc}\frac{a}{\sqrt{2b^2+2c^2-a^2}}\geq\sqrt3.$$ Indeed, by Holder $$\left(\sum_{cyc}\frac{a}{\sqrt{2b^2+2c^2-a^2}}\right)^2\sum_{cyc}a(2b^2+2c^2-a^2)\geq(a+b+c)^3.$$ Thus, it remains to prove that $$(a+b+c)^3\geq3\sum_{cyc}a(2b^2+2c^2-a^2)$$ or $$\sum_{cyc}(4a^3-3a^2b-3a^2c+2abc)\geq0$$ or $$2\sum_{cyc}(a^3-a^2b-a^2c+abc)+\sum_{cyc}(2a^3-a^2b-a^2c)\geq0,$$ which is Schur and Muirhead.

Done!

There is also solution by C-S.

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