2. Given $a^{2}+b^{2}+c^{2}=1$, then the range of $a b+b c+a c$ is
2. $\left[-\frac{1}{2}, 1\right]$

Analysis: Clearly, $a b+b c+c a \leq a^{2}+b^{2}+c^{2}=1$, with equality holding if and only if $a=b=c$. On the other hand, $a b+b c+c a=a(b+c)+b c \geq-\frac{a^{2}+(b+c)^{2}}{2}+b c=-\frac{a^{2}+b^{2}+c^{2}}{2}=-\frac{1}{2}$, with equality when $a+b+c=0$, so the range is $\left[-\frac{1}{2}, 1\right]$