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? 
 A: 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 $a=b=c=1$.
A: The following substitution transforms the constrained inequality into an unconstrained one:
$$a = t^2 \gt 0$$
$$b = u^2 \gt 0$$
$$c = v^2 \gt 0$$
The original inequality becomes:
$$F(t, u, v) = t^4+u^4+v^4 +2t^2u^2v^2+1 - 2*(t^2u^2 + t^2v^2 + u^2v^2) \ge 0$$
The new expression is fully symmetric in $t, u $ and $v$.
Therefore, the minimum must be symmetric as well, which means $t=u=v$.
$$F(t,t,t) = -3t^4 + 2t^6 + 1$$
$$\frac{\delta F}{\delta t} = -12t^33 + 12 t^5 = 12t^3(t^2-1)$$
For $t=1$ we get an extremum $F(1,1,1)=0$ which confirms our original inequality.

The following steps should convince us that the solution is in fact symmetric:
To be at the minimum of the left-hand-side, the partial derivatives have to be zero.
$$\frac{\delta F}{\delta t} = 4t^3 + 4tu^2v^2 - 4tu^2 - 4tv^2 = 0$$
$$\frac{\delta F}{\delta u} = 4u^3 + 4t^2uv^2 - 4t^2u - 4uv^2 = 0$$
$$\frac{\delta F}{\delta v} = 4v^3 + 4t^2u^2v - 4t^2v - 4u^2v = 0$$
this leads to:
$$t(t^2 - v^2 + u^2(v^2 - 1)) = 0$$
$$u(u^2 - v^2 + t^2(v^2 - 1)) = 0$$
$$v(v^2 - u^2 + t^2(u^2 - 1)) = 0$$
All variables have to be greater than zero.
$$t^2 - v^2 + u^2(v^2 - 1) = 0$$
$$u^2 - v^2 + t^2(v^2 - 1) = 0$$
$$v^2 - u^2 + t^2(u^2 - 1) = 0$$
The first two of these equations combine to:
$$ (t^2 - u^2)v^2 = 0$$
The last two equations give us:
$$ t^2(u^2+v^2-2) = 0$$
The first and the third equation yield:
$$t^2 - u^2 +u^2(v^2-1) + t^2(u^2-1)$$
Assuming $t=u$ we get:
$$t^2(t^2+v^2-2) = 0$$
or
$$v^2 = 2 - t^2$$
Inserting this we get
$$- t^4 + 3t^2 -2 = 0$$
This can be factored into
$$(t-1)(t+1)(t^2-2)=0$$
For $t^2-2=0$ we get $a=2$, $b=2$ and $c=0$ which violates the constraints.
Therefore, $a=b=c=1$ is the global minmum.
A: Let $a=x^3$, $b=y^3$ and $c=z^3$.
Hence, by AM-GM and Schur we obtain:
$$a^2+b^2+c^2+2abc+1\geq a^2+b^2+c^2+3\sqrt{a^2b^2c^2}=\sum_{cyc}(x^6+x^2y^2z^2)\geq$$
$$\geq\sum_{cyc}(x^4y^2+y^4x^2)\geq2\sum_{cyc}x^3y^3=2(ab+ac+bc)$$
and we are done!
