Let $a,b,c$ be the side lengths of any triangle. Prove that $$\frac{a}{\sqrt{2b^2+2c^2-a^2}}+\frac{b}{\sqrt{2c^2+2a^2-b^2 }}+\frac{c}{\sqrt{2a^2+2b^2-c^2}}\ge \sqrt{3}.$$(Zhuge Liang)
If we multiply everything by $2$, we can get
$\sum_{cyc} \frac{a}{m_a} \ge 2\sqrt{3}$ with $m_a$ stands for the median to the side $a$. Now, the question is, does this help xd
parmenides51 wrote:
Let $a,b,c$ be the side lengths of any triangle. Prove that $$\frac{a}{\sqrt{2b^2+2c^2-a^2}}+\frac{b}{\sqrt{2c^2+2a^2-b^2 }}+\frac{c}{\sqrt{2a^2+2b^2-c^2}}\ge \sqrt{3}.$$(Zhuge Liang)
By AM-GM $$\sum_{cyc}\frac{a}{\sqrt{2b^2+2c^2-a^2}}=\sum_{cyc}\frac{2\sqrt3a^2}{2\sqrt{3a^2(2b^2+2c^2-a^2)}}\geq\sum_{cyc}\frac{2\sqrt3a^2}{2(a^2+b^2+c^2)}=\sqrt3.$$
#3 do you think using medians would lead to anywhere? Is there any known inequality with medians of which you are familiar?
BTW, awesome solution with AM-GM