Ta có : \(a^2+b^2+1\ge ab+a+b\)
\(\Leftrightarrow\) \(2a^2+2b^2+2\ge2ab+2a+2b\)
\(\Leftrightarrow\) \(2a^2+2b^2+2-2ab-2a-2b\ge0\)
\(\Leftrightarrow\) \(\left(a^2-2ab+b^2\right)+\left(a^2-2a+1\right)+\left(b^2-2b+1\right)\ge0\)
\(\Leftrightarrow\) \(\left(a-b\right)^2+\left(a-1\right)^2+\left(b-1\right)^2\ge0\) ( Luôn đúng )
\(\Rightarrow\) \(a^2+b^2+1\ge ab+a+b\)
Ta có :a2+b2+1≥ab+a+b
⇔2a2+2b2+2≥ 2ab+2a+2b
⇔2a2+2b2+2-2ab-2a-2b≥0
⇔(a2-2ab+b2)+(a2-2a+1)+(b2-2b+1)≥0
⇔(a-b)2+(a-1)2+(b-1)2≥0 (1)
Vì (a-b)≥0, ∀a,b.(2)
(a-1)2≥0, ∀a (3) (luôn đúng)
(b-1)2≥0, ∀b (4)
Từ (1), (2),(3),(4) ⇒ \(\left[{}\begin{matrix}\left(a-b\right)^2=0\\\left(a-1\right)^2=0\\\left(b-1\right)^2=0\end{matrix}\right.\)⇔\(\left[{}\begin{matrix}a=b\\a=1\\b=1\end{matrix}\right.\)
Vậy a2+b2+1≥ab+a+b khi a=b=1