\(a^2+b^2+1\ge ab+a+b\)
\(<=>2a^2+2b^2+2\geq 2ab+2a+2b\\<=>(a^2-2ab+b^2)+(a^2-2a+1)+(b^2-2b+1)\geq 0\\<=>(a-b)^2+(a-1)^2+(b-1)^2\geq 0\)
$\Rightarrow $ \(a^{2}+b^{2}\geq 2ab\) (1)
$\Rightarrow $ \(a^{2}+1\geq 2a\) (2)
$\Rightarrow $ \(b^{2}+1\geq 2b\) (3)
(1), (2) và (3)\(\Rightarrow a^{2}+b^{2}+1\geq ab+a+b\)