a, Ta có: \(\left(a+b\right)^2\ge0\Rightarrow a^2+b^2\ge2ab\)
\(\left(a+1\right)^2\ge0\Rightarrow a^2+1\ge2a\)
\(\left(b+1\right)^2\ge0\Rightarrow b^2+1\ge2b\)
Cộng vế với vế ta có: \(a^2+b^2+a^2+1+b^2+1\ge2ab+2a+2b\)
\(\Rightarrow2a^2+2b^2+2\ge2ab+2a+2b\)
\(\Rightarrow2\left(a^2+b^2+1\right)\ge2\left(ab+a+b\right)\)
\(\Rightarrow a^2+b^2+1\ge ab+a+b\)(ĐPCM)