jmc

By Cauchy-Schwarz, $$(a^2+2b^2+c^2)\left(1+\frac{1}{2}+1\right)\ge (a+b+c{)}^2=1,$$so $a^2+2b^2+c^2\ge \frac{2}{5}.$

Equality occurs when $\frac{a^2}{1}=\frac{2b^2}{1/2}=\frac{c^2}{1}$ and $a+b+c=1.$ We can solve to get $a=\frac{2}{5},$ $b=\frac{1}{5},$ and $c=\frac{2}{5},$ so the minimum value is $\boxed{\frac{2}{5}}.$