jmc

By QM-AM, we have $$\sqrt{\frac{(1-x_1{)}^2+(x_1-x_2{)}^2+(x_2-x_3{)}^2+x_3^2}{4}}\ge \frac{(1-x_1)+(x_1-x_2)+(x_2-x_3)+x_3}{4}=\frac{1}{4}.$$Taking the square of both sides, and then multiplying both sides by $4$ gives us, $$(1-x_1{)}^2+(x_1-x_2{)}^2+(x_2-x_3{)}^2+x_3^2\ge \frac{1}{4}.$$Equality occurs if and only if $1-x_1=x_1-x_2=x_2-x_3=x_3=\frac{1}{4}$. We can solve to get $x_1=\boxed{\frac{3}{4}},$ $x_2=\frac{1}{2},$ and $x_3=\frac{1}{4}.$