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? 

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.
