<s>[INST] True or false: $a^2+b^2+c^2 +2abc+1\geq 2(ab+bc+ca)$ Is this inequality true?

$a^2+b^2+c^2 +2abc+1\ge2(ab+bc+ca)$, where $a,b,c\gt0$.

Can you find a counterexample for this or not? 
 [/INST] EDIT: I found a much better solution:
WLOG assume that $b,c$ are on the same side of $1$. Bringing the RHS over to the LHS, it suffices to show that
$$(a-1)^2+2a(b-1)(c-1)+(b-c)^2\ge 0$$
But since $b,c$ are on the same side of $1$, this is true.

Old solution:
We prove the inequality for $a,b,c\ge 0$.
The inequality is equivalently
$$
a^2+2(bc-b-c)a+b^2+c^2+1-2bc\ge 0
$$
If $bc>b+c$, then the expression is minimized when $a=0$. But in this case it is only left to prove $b^2+c^2-2bc+1=(b-c)^2+1\ge 0$, which is clear.
If $bc\le b+c$, then then expression is minimized when $a = b+c-bc$. We must show
$$
-(bc-b-c)^2+b^2+c^2+1\ge 2bc\Leftrightarrow 2(b+c)bc+1\ge 4bc+(bc)^2
$$
But letting $bc=k$, we have
$$
2(b+c)bc+1\ge 4k^{3/2}+1
$$
And so we want to prove
$$
k^2-4k^{3/2}+4k-1=(\sqrt{k}-1)^2(k-2\sqrt{k}-1)\le 0
$$
given $bc\le b+c\le 2\sqrt{bc}\Rightarrow 0\le k\le 4$. But then $k-2\sqrt{k}-1=\sqrt{k}(\sqrt{k}-2)-1\le -1<0$, so the inequality holds.
Equality holds when $a=b=c=1$.
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