, 2019

Let $a + b + c = ab + bc + ca = k$. Since $(a + b + c)^2 \ge 3(ab + bc + ca)$, we get that $k^2 \ge 3k$. Since $k > 0$, we obtain that $k \ge 3$.
We have $bc \ge ca \ge ab$, so from the above relation we deduce that $bc \ge 1$.
By AM-GM, $b + c \ge 2\sqrt{bc}$ and consequently $b + c \ge 2$. The equality holds iff $b = c$.
The constraint gives us
$$a = \frac{b + c - bc}{b + c - 1} = 1 - \frac{bc - 1}{b + c - 1} \ge 1 - \frac{bc - 1}{2\sqrt{bc} - 1} = \frac{\sqrt{bc}(2 - \sqrt{bc})}{2\sqrt{bc} - 1}.$$
For $\sqrt{bc} = 2$ condition $a \ge 0$ gives $\sqrt{bc}(a + 1) \ge 2$ with equality iff $a = 0$ and $b = c = 2$.
For $\sqrt{bc} < 2$, taking into account the estimation for $a$, we get
$$a\sqrt{bc} \ge \frac{bc(2 - \sqrt{bc})}{2\sqrt{bc} - 1} = \frac{bc}{2\sqrt{bc} - 1}(2 - \sqrt{bc}).$$
Since $\frac{bc}{2\sqrt{bc} - 1} \ge 1$, with equality for $bc = 1$, we get $\sqrt{bc}(a + 1) \ge 2$ with equality iff $a = b = c = 1$.
For $\sqrt{bc} > 2$ we have $\sqrt{bc}(a + 1) > 2(a + 1) \ge 2$.
The proof is complete.
The equality holds iff $a = b = c = 1$ or $a = 0$ and $b = c = 2$. $\square$