Don't make it complicated

Rewrite the expression $(1+a+b+c)^2-(ab+bc+ca)$ $\Leftrightarrow 1+2(a+b+c)+a^2+b^2+c^2+ab+bc+ac$ Then just apply AM-GM to $a+b+c$ ; $a^2+b^2+c^2$ ; $ab+bc+ac$ and we get the answer $13$

Noticing that a b c can be the roots of x³+Px²+Qx-1=0,where a+b+c=-P ab+bc+ac=Q(P＜0,Q＞0)，（abc+a+b+c）²-（ab+bc+ac）=（1-P）²-Q

since （a+b+c）²≥3（ab+bc+ac）,P²≥3Q.

considering that a+b+c≥3(abc)^(1/3)=3, P≤-3.

hence,（1-P）²-Q≥1/3（-P²）+P²-2P+1=2/3（P²）-2P+1≥13

so the answer is 13（P=-3 Q=3）