jmc

In the coordinate plane, let $A=(0,1),$ $B=(1,-1),$ and $P=(x,x).$ Then $$AP=\sqrt{x^2+(1-x{)}^2}$$and $$BP=\sqrt{(x-1{)}^2+(x+1{)}^2},$$so we want to minimize $AP+BP,$ subject to $P$ lying on the line $y=x.$



By the Triangle Inequality, $AP+BP\ge AB=\sqrt{5}.$ Equality occurs when $P$ is the intersection of the line $y=x$ and line $AB$ (which occurs when $x=\frac{1}{3}$), so the minimum value is $\boxed{\sqrt{5}}.$