a/ \(f\left(x\right)=2x^2-2\left(a+b\right)x+a^2+b^2\)
\(=\frac{1}{2}\left[4x^2-4\left(a+b\right)x+\left(a+b\right)^2\right]+\frac{1}{2}a^2+\frac{1}{2}b^2-ab\)
\(=\frac{1}{2}\left(2x-a-b\right)^2+\frac{1}{2}\left(a-b\right)^2\ge\frac{1}{2}\left(a-b\right)^2\)
Dấu "=" xảy ra khi \(x=\frac{a+b}{2}\)
b/ \(f\left(x\right)=3x^2-2\left(a+b+c\right)x+a^2+b^2+c^2\)
\(=\frac{1}{3}\left[9x^2-6\left(a+b+c\right)x+\left(a+b+c\right)^2\right]+\frac{2}{3}\left(a^2+b^2+c^2\right)-\frac{2}{3}\left(ab+bc+ca\right)\)
\(=\frac{1}{3}\left(3a-a-b-c\right)^2+\frac{2}{3}\left(a^2+b^2+c^2-ab-bc-ca\right)\)
\(\Rightarrow f\left(x\right)_{min}=\frac{2}{3}\left(a^2+b^2+c^2-ab-bc-ca\right)\) khi \(x=\frac{a+b+c}{3}\)
