Prove that there is no triple $(a,b,c)$ Prove that there is no triple $(a,b,c)$ such that
$$a,b,c \geq 0,ab+bc+ca=2+abc \textrm{ and } abc \in (1;10+6\sqrt{3}).$$
I transform the expression as follows
$$ab = \frac{m}{c},a + b = \frac{{2 + m - \frac{m}{c}}}{c}.$$
I expect the following inequality to be false and thus conclude that there is no satisfying set of numbers
$${\left( {a + b} \right)^2} \ge 4ab \Leftrightarrow {\left( {\frac{{2 + m - \frac{m}{c}}}{c}} \right)^2} - 4\frac{m}{c} \ge 0.$$
But it's quite complicated

$a,b,c \geq 0, \; ab+bc+ca=2+abc, \; abc \in (1,10+6\sqrt{3})$
By AM-GM inequality, $\frac{ab+bc+ca}{3} \geq (abc)^{\frac{2}{3}} \implies ab+bc+ca \geq 3(abc)^{\frac{2}{3}} \implies 2+abc \geq 3(abc)^{\frac{2}{3}}$
Consider $abc = x^3$, then $x^3-3x^2+2 \geq 0 \implies (x-1)(x-1+\sqrt{3})(x-1-\sqrt{3}) \geq 0$
It can be easily checked that this is only possible when $x \in [1-\sqrt{3},1] \cup [1+\sqrt{3}, \infty)$
So, $abc = x^3 \in [10-6\sqrt{3},1] \cup [10+6\sqrt{3}, \infty)$
Therefore, there is no such triple (a,b,c) such that $abc \in (1, 10+6\sqrt{3})$
