\(a^2+b^2+1\ge ab+a+b\)
\(\Leftrightarrow2a^2+2b^2+2-2ab-2a-2b\ge0\)
\(\Leftrightarrow\left(a^2-2ab+b^2\right)+\left(a^2-2a+1\right)+\left(b^2-2b+1\right)\ge0\)
\(\Leftrightarrow\left(a-b\right)^2+\left(a-1\right)^2+\left(b-1\right)^2\ge0\left(true!!\right)\)
Dấu "=" xảy ra tại a=b=1
Xét hiệu \(A=\left(a^2+b^2+1\right)-\left(ab+a+b\right)\)
\(=a^2+b^2+1-ab-a-b\)
\(\Rightarrow2A=2a^2+2b^2+2-2ab-2a-2b\)
\(=\left(a^2-2ab+b^2\right)+\left(a^2-2a+1\right)+\left(b^2-2b+1\right)\)
\(=\left(a-b\right)^2+\left(a-1\right)^2+\left(b-1\right)^2\ge0\)
\(\Rightarrow2A\ge0\Leftrightarrow A\ge0\)
Vậy \(a^2+b^2+1\ge ab+a+b\left(đpcm\right)\)
Dấu " = " xảy ra \(\Leftrightarrow\hept{\begin{cases}a-b=0\\a-1=0\\b-1=0\end{cases}}\Leftrightarrow a=b=1\)
Xét hiệu: \(VT-VP=\frac{3}{4}\left(a-b\right)^2+\frac{1}{4}\left(a+b-2\right)^2\ge0\)
Đẳng thức xảy ra khi \(a=b=1\)