Yes, it reaches a minimal value of $1$ when $x=0$.

$\underline{\textbf{Solution 1}}$ Because $(x-1)^2 -(2x^2 -2x+1) = -x^2 \le 0$, we have $$(x-1)^2 \le (2x^2-2x+1)\implies 1-x\le|1-x|\le\sqrt{2x^2-2x+1}$$ Therefore, $$1\le x+\sqrt{2x^2-2x+1}$$ The equality holds when $x=0$.

$\underline{\textbf{Solution 2}}$ Constructing a right triangle whose two sides are $|x|$ and $|x-1|$. respectively. Then its hypotenuse is $$\sqrt{x^2+(x-1)^2}=\sqrt{2x^2-2x+1}$$ Clearly, hypotenuse should be no shorter than the side, therefore $$\sqrt{2x^2-2x+1}\ge |x-1|\ge 1-x \implies x+\sqrt{2x^2-2x+1}\ge 1$$