Prove $\frac{1}{\sqrt{4a+4b+1}}+\frac{1}{\sqrt{4c+4b+1}}+\frac{1}{\sqrt{4a+4c+1}}\ge 1,$ for $ab+bc+ca=abc+2$

Question

Let $a,b,c\ge 0: ab+bc+ca=abc+2.$ Prove that $$\frac{1}{\sqrt{4a+4b+1}}+\frac{1}{\sqrt{4c+4b+1}}+\frac{1}{\sqrt{4a+4c+1}}\ge 1.$$

I tried to use C-S $$\sum_{cyc}\frac{1}{\sqrt{4a+4b+1}}\ge \frac{9}{\sum_{cyc}\sqrt{4a+4b+1}}\ge 1.$$ Now, we need to prove $$\sqrt{4a+4b+1}+\sqrt{4c+4b+1}+\sqrt{4a+4c+1}\le 9.$$ It is not true. I think Holder might be useful but I couldn't find it yet.

Hope to see some ideas. I'm appreciate your interest.

Update.

I just see that equality occurs when $a=b=c=1$ also for $a=b=0$

Answer 1

It's very, very hard to prove --- because it is false.

$b$ and $c$ can become arbitrarily large as long as $a$ remains close enough to $1$, and in this case the sum can be arbitrarily close to $0$.

Conversely, if $b = c$ are arbitrarily close to zero, $a$ becomes arbitrarily large but the sum still remains greater than $1$, so no ball centered on the origin contains all of the feasible set.