BMO 2019 Shortlist

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 $b=c=2$. $\square$