For $a,b,c>0$ and $a+b+c=3$. Prove that$:$ $$\sqrt{3a+\frac{1}{b}} +\sqrt{3b+\frac{1}{c}} +\sqrt{3c+\frac{1}{a}} \geqq 6$$
My work$:$
This inequality equivalent to$:$
$$3(a+b+c) +\frac{1}{a}+\frac{1}{b}+\frac{1}{c} +2\sum \sqrt{(3a+\frac{1}{b})(3b+\frac{1}{c})}\geqq 36$$
Or $$ \frac{1}{a}+\frac{1}{b}+\frac{1}{c} +2\sum \sqrt{(3a+\frac{1}{b})(3b+\frac{1}{c})} \geqq 27$$
By C-S$,$ $$\sqrt{(2a+a+\frac{1}{b})(2b+\frac{1}{c}+b)} \geqq (2\sqrt{ab} +\sqrt{\frac{a}{c}}+1)$$
Hence$,$ it is enough to prove: $$\frac{1}{a}+\frac{1}{b}+\frac{1}{c} +2 \sum (2\sqrt{ab} +\sqrt{\frac{a}{c}}) \geqq 21$$
But by AM-GM$,$ $$\frac{1}{a} +\frac{1}{b}+\frac{1}{c} +\sum \sqrt{ab} \geqq 3 \sqrt[3]{\frac{1}{abc}} +3\sqrt[3]{abc} \geqq 6$$
So we need to prove$:$ $$3 \sum \sqrt{ab} +2\sum \sqrt{\frac{a}{c}} \geqq 15$$
But I don't know what I need to do next$?$
