Let $a+b+c=ab+bc+ca=k$. Since $(a+b+c)^{2} \geq 3(ab+bc+ca)$, we get that $k^{2} \geq 3k$. Since $k>0$, we obtain that $k \geq 3$.

We have $bc \geq ca \geq ab$, so from the above relation we deduce that $bc \geq 1$.
By AM-GM, $b+c \geq 2 \sqrt{bc}$ and consequently $b+c \geq 2$. The equality holds iff $b=c$.
The constraint gives us

$$a=\frac{b+c-bc}{b+c-1}=1-\frac{bc-1}{b+c-1} \geq 1-\frac{bc-1}{2 \sqrt{bc}-1}=\frac{\sqrt{bc}(2-\sqrt{bc})}{2 \sqrt{bc}-1}$$

For $\sqrt{bc}=2$ condition $a \geq 0$ gives $\sqrt{bc}(a+1) \geq 2$ with equality iff $a=0$ and $b=c=2$.
For $\sqrt{bc}>2$ we have $\sqrt{bc}(a+1)>2(a+1) \geq 2$.
The proof is complete.
The equality holds iff $a=b=c=1$ or $a=0$ and $b=c=2$.