\(a^2+b^2+1=ab+a+b\)
\(\Leftrightarrow2\left(a^2+b^2+1\right)=2\left(ab+a+b\right)\)
\(\Leftrightarrow2a^2+2b^2+2=2ab+2a+2b\)
\(\Leftrightarrow2a^2+2b^2+2-2ab-2a-2b=0\)
\(\Leftrightarrow\left(a^2-2ab+b^2\right)+\left(a^2-2a+1\right)+\left(b^2-2b+1\right)=0\)
\(\Leftrightarrow\left(a-b\right)^2+\left(a-1\right)^2+\left(b-1\right)^2=0\)
Do \(\left(a-b\right)^2\ge0;\left(a-1\right)^2\ge0;\left(b-1\right)^2\ge0\)
\(\Rightarrow\left(a-b\right)^2+\left(a-1\right)^2+\left(b-1\right)^2\ge0\)
Dấu " = " xảy ra
\(\Leftrightarrow\left\{{}\begin{matrix}a-b=0\\a-1=0\\b-1=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a=b\\a=1\\b=1\end{matrix}\right.\)
Vậy \(a=b=1\)
