\(2a^2-2a+1\ge\dfrac{1}{2}\)
\(\Leftrightarrow2\left(2a^2-2a+1\right)\ge2\cdot\dfrac{1}{2}\)(nhân 2 vế cho 2>0)
\(\Leftrightarrow4a^2-4a+2\ge1\)
\(\Leftrightarrow4a^2-4a+1\ge0\)(chuyển vế)
\(\Leftrightarrow\left(2a-1\right)^2\ge0\)(luôn đúng)
Vậy \(S=R\)