How to prove $\sqrt{\frac{ab}{2a^2+bc+ca}}+\sqrt{\frac{bc}{2b^2+ca+ab}}+\sqrt{\frac{ca}{2c^2+ab+bc}}\ge\frac{81}{2}\cdot\frac{abc}{(a+b+c)^3}$

Question

Question:

let $a,b,c>0$,show that:

$$\sqrt{\frac{ab}{2a^2+bc+ca}}+\sqrt{\frac{bc}{2b^2+ca+ab}}+\sqrt{\frac{ca}{2c^2+ab+bc}}\ge\frac{81}{2}\cdot\frac{abc}{(a+b+c)^3}$$

maybe this inequality can Use Holder inequality to solve it $$\left(\sum_{cyc}\sqrt{\dfrac{ab}{2a^2+bc+ca}}\right)^2\sum_{cyc}(2a^2+bc+ca)\ge(\sum_{cyc}\sqrt[3]{ab})^3$$

then I can't prove it

Answer 1

By Hölder's inequality we have $$\left(\sum_{cyc}\sqrt{\frac{a}{2bc + ca + abc}}\right)^2 \cdot \left(\sum_{cyc}\frac{2bc + ca + abc}{a}\right) \geq 27$$ Hence it suffices to prove that $$27 \geq \frac{81}{2} \cdot \frac{abc}{(a + b + c)^3} \cdot \left(\sum_{cyc}\frac{2bc + ca + abc}{a}\right)$$ which is obvious after homogenization and expansion