<s>[INST] Prove $\frac{1}{2a+2bc+1} + \frac{1}{2b+2ca+1} + \frac{1}{2c+2ab+1} \ge 1$ If $a,b$ and $c \ge 0$ and $ab + bc + ca = 1$, prove that the following inequality holds:
$$\frac{1}{2a+2bc+1} + \frac{1}{2b+2ca+1} + \frac{1}{2c+2ab+1} \ge 1$$

I've tried two aproaches, but it seems like both doesn't work. Here they are:
Cauchy-Scwarz inequality
I've tried using the following formula of Cauchy-Scwarz:
$$\frac{x_1^2}{a_1} + \frac{x_2^2}{a_2} + \frac{x_3^2}{a_3} \ge \frac{(x_1 + x_2 + x_3)^2}{a_1 + a_2 + a_3}$$
And i get:
$$ LHS \ge \frac{(1+1+1)^2}{2a + 2bc + 2b + 2ca + 2c + 2ba + 3}$$
$$ LHS \ge \frac{3^2}{2(a+b+c) + 5}$$
$$ LHS \ge \frac{9}{2(a+b+c) + 5}$$
Now to prove that the RHS is bigger than or equal to 1.
$$ \frac{9}{2(a+b+c) + 5} \ge 1$$
$$ 9 \ge 2(a+b+c) + 5$$
$$ 2 \ge a+b+c$$
And her I'm stuck, what can I do now?
AM - GM inequality
This try doesn't even stand a chance, but I'll still post something, because somebody can recieve an idea.
$$2a + 2bc \ge 2\sqrt{4abc}$$
$$2a + 2bc + 1 \ge 4\sqrt{abc} + 1$$
$$\frac{1}{2a + 2bc + 1} \le \frac{1}{4\sqrt{abc} + 1}$$
And for oher 2 fraction i get the same and if I add them I get:
$$\frac{3}{4\sqrt{abc} + 1} \ge LHS \ge 1$$
Now even if i prove that: $\frac{3}{4\sqrt{abc} + 1} \ge 1$ si true, that doesn't mean the original inequality holds.
 [/INST] Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Hence, our inequality it's $f(w^3)\geq0$, where $f(w^3)=-8w^6+A(u,v^2)w^3+B(u,v^2)$.
But $f$ is a concave function, which says that it's enough to prove our inequality for an extremal value of $w^3$ wich happens in the following cases.


*

*$w^3=0$.


Let $c=0$. Hence, $ab=1$ and we need to prove that
$$\frac{1}{2a+1}+\frac{1}{2b+1}+\frac{1}{3}\geq1$$ or
$$\frac{a+b+1}{2(a+b)+5}\geq\frac{1}{3}$$ or
$$a+b\geq2,$$
which is AM-GM: $a+b\geq2\sqrt{ab}=2$;


*Two variables are equal.


Let $b=a$. Hence, $a\neq0$ and $c=\frac{1-a^2}{2a}$, where $0<a<1$ 
and after these substitutions we need to prove that
$$a^2(1-a)(1+3a-2a^2)\geq0.$$
Done!
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