Prove: $\frac{a}{m_a}+\frac{b}{m_b}+\frac{c}{m_c}\ge 2\sqrt{3}$

Question

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.

Answer 1

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!

Answer 2

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.

Answer 3

Another way.

By the Ptolemy and C-S twice we obtain: $$\sum_{cyc}\frac{a}{m_a}=\sqrt{\sum_{cyc}\left(\frac{a^2}{m_a^2}+\frac{2ab}{m_am_b}\right)}\geq\sqrt{\sum_{cyc}\left(\frac{a^2}{m_a^2}+\frac{2ab}{\frac{c^2}{2}+\frac{ab}{4}}\right)}=$$ $$=2\sqrt{\sum_{cyc}\left(\frac{a^2}{2b^2+2c^2-a^2}+\frac{2ab}{ab+2c^2}\right)}=2\sqrt{\sum_{cyc}\left(\frac{a^4}{2a^2b^2+2a^2c^2-a^4}+\frac{2a^2b^2}{a^2b^2+2c^2ab}\right)}\geq$$ $$\geq2\sqrt{\frac{(a^2+b^2+c^2)^2}{\sum\limits_{cyc}(4a^2b^2-a^4)}+\frac{2(ab+ac+bc)^2}{\sum\limits_{cyc}(a^2b^2+2c^2ab)}}\geq2\sqrt{1+2}=2\sqrt3.$$