True or false: $a^2+b^2+c^2 +2abc+1\geq 2(ab+bc+ca)$

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.

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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 $a=b=c=1$.