Xét hiệu \(T=a^2+b^2+1-\left(ab+a+b\right)\ge0\)
\(=a^2+b^2+1-ab-a-b\ge0\)
Ta có: \(2T=2a^2+2b^2+2-2ab+2a-2b\ge0\)
\(=\left(a^2-2ab+b^2\right)+\left(a^2-2a+1\right)+\left(b^2-2b+1\right)\ge0\)
\(=\left(a-b\right)^2+\left(a-1\right)^2+\left(b-1\right)^2\ge0\)
(ĐPCM)
Dấu = khi \(\begin{cases}b-1=0\\a-1-0\\a-b=0\end{cases}\)\(\Leftrightarrow a=b=1\)