If $a,b,c>0 : a+b+c=2(ab+bc+ca),$ then find the minimal value$$P=\sqrt{\dfrac{1}{ab}+\dfrac{1}{bc}+1}+\sqrt{\dfrac{1}{bc}+\dfrac{1}{ca}+1}+\sqrt{\dfrac{1}{ca}+\dfrac{1}{ab}+1}.$$
By set $a=b=c=1/2,$ I got $P\ge 9.$
I've tried to use classical inequality based on this equality case. For example, $$4.\frac{1}{4ab}+4.\frac{1}{4cb}+1\ge 9\sqrt[9]{\frac{1}{(4ab)^4.(4bc)^4}}$$
Similarly, it's enough to prove $$\sum_{cyc}\sqrt[18]{\frac{1}{(4ab)^4.(4bc)^4}}\ge 3$$
Also by AM-GM, it remains to prove $abc\le \dfrac{1}{8}.$
I can't go further because I am still stuck to show $abc\le \dfrac{1}{8}$ is true or false.
Hope you can help me continue my idea. Thank you.
Best Answer
Proof.
By your idea to prove $P\ge 9,$ we can apply AM-GM inequality as $$2\sqrt{a+b+abc}\le \sqrt{\frac{ab}{c}}(c+1)+\frac{a+b+abc}{c+1}\sqrt{\frac{c}{ab}}=\frac{ab+bc+ca}{(c+1)\sqrt{abc}}+2\sqrt{abc}.$$ It implies that $$\frac{ab+bc+ca}{(c+1)\sqrt{abc}}\ge \frac{2(a+b)}{\sqrt{a+b+abc}+\sqrt{abc}},$$or $$\sqrt{\dfrac{1}{ab}+\dfrac{1}{bc}+1} \ge \dfrac{2a+2b+ca+cb-ab}{ab+bc+ca} \tag{1}.$$ Take cyclic sum on $(1),$ we get \begin{align*} &\sqrt{\dfrac{1}{ab}+\dfrac{1}{bc}+1}+\sqrt{\dfrac{1}{bc}+\dfrac{1}{ca}+1}+\sqrt{\dfrac{1}{ca}+\dfrac{1}{ab}+1}\\ &\ge \dfrac{4(a+b+c)+ab+bc+ca}{ab+bc+ca}\\ &\ge \dfrac{9(ab+bc+ca)}{ab+bc+ca}=9 \end{align*} Hence, the proof is done.
In conclusion: Minimum $P=9$ at $a=b=c=\dfrac{1}{2}.$