ĐK: $x\ge0$.
Ta có:
$2(\sqrt x-1)^2\ge 0 \Leftrightarrow 1-4\sqrt x\ge-2x-1 \Leftrightarrow \dfrac{1-4\sqrt x}{2x+1}\ge-1$
$(x-1)^2\ge0 \Leftrightarrow -2x\ge-x^2-1 \Leftrightarrow -\dfrac{2x}{x^2+1}\ge-1$
Từ đó: $S\ge-2$.
Vậy $\min S=-2 \Leftrightarrow x=1$