jmc

By the Trivial Inequality, $(a+b+c{)}^2\ge 0.$ This expands as $$a^2+b^2+c^2+2ab+2ac+2bc\ge 0.$$We know $a^2+b^2+c^2=1,$ so $$2ab+2ac+2bc+1\ge 0.$$Hence, $$ab+ac+bc\ge -\frac{1}{2}.$$Equality occurs when $a=0,$ $b=\frac{1}{\sqrt{2}},$ and $c=-\frac{1}{\sqrt{2}}.$ Therefore, the minimum value of $ab+ac+bc$ is $\boxed{-\frac{1}{2}}.$