Ta có: $0\leq (a+b+c)^2=(a^2+b^2+c^2)+2(ab+bc+ca)$$\Rightarrow ab+bc+ca\geq -\frac{1}{2}$
Đẳng thức khi $\left\{ \begin{array}{l} a+b+c=0\\ a^2+b^2+c^2=1 \end{array} \right.$
Mặt khác : $\frac{a^2+b^2}{2}$$\geq$ $-ab\Rightarrow$ $ab\geq$ -$\frac{a^2+b^2}{2}$
$\Rightarrow F=ab+bc+ca+bc\geq -4-\frac{a^2+c^2}{2}$
Ta có : $a^2+b^2\leq a^2+b^2+c^2=1$
$\Rightarrow \frac{1}{2}(a^2+c^2)\geq -\frac{1}{2}\Rightarrow F\geq -1$
Vậy min $F=-1$ khi $\left\{ \begin{array}{l} c=0\\ a+b=0 \end{array} \right.$
$\Rightarrow ...$