Problem. Let $a,b,c\ge 0:ab+bc+ca=1.$ Prove that $$\frac{1}{\sqrt{a+bc}}+\frac{1}{\sqrt{b+ca}}+\frac{1}{\sqrt{c+ab}}\ge 2\sqrt{2}.$$ Equality holds when two of $a,b,c$ very near $0$.
The form of problem is quite similar with Pham Kim Hung's inequality (AOPSlink) and a problem I have posted recently (similarMSE)
In the post #46, Michael Rozenberg (aka arqady) gave a very sharp Holder estimate.
They are two same equality case. That why I tried to use similar Holder approach: \begin{align*} \left(\sum_{cyc}\frac{1}{\sqrt{a+bc}}\right)^2\sum_{cyc}(a+bc)(b+c)^3(3a+b+c)^3&\geq\left(\sum_{cyc}(b+c)(3a+b+c)\right)^3\\&=8\left(\sum_{cyc}(a^2+4ab)\right)^3 \end{align*}Hence, we need to prove:$$\left(\sum_{cyc}(a^2+4ab)\right)^3\geq\sum_{cyc}(a+bc)(b+c)^3(3a+b+c)^3,$$hay$$\left((a+b+c)^2+2\right)^3$$ $$ \geq (a+bc)(b+c)^3(3a+b+c)^3+(b+ca)(a+c)^3(3b+a+c)^3+(c+ba)(b+a)^3(3c+b+a)^3,$$ which is wrong when $a=b\rightarrow 0;c\rightarrow +\infty !$
I am trying to find a good Holder estimate.
If you find something interesting in using Holder (like tricks, the motivations...) please share me. Thank you.