Question

CM: \(a^2+b^2+1\ge ab+a+b\)

Answer 1

Giải:

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 \(a^2+b^2+1\ge ab+a+b\) (Đpcm)

Dấu "=" xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}a-b=0\\a-1=0\\b-1=0\end{matrix}\right.\Leftrightarrow a=b=1\)