Let a,b be real nmubers.Prove that \(P=a^2+b^2-a-b-ab\ge-1\)
Luân Đào\(a^2+b^2-a-b-ab\ge-1\)
\(\Leftrightarrow2a^2+2b^2-2a-2b-2ab+2\ge0\)
\(\Leftrightarrow\left(a-b\right)^2+\left(a-1\right)^2+\left(a-1\right)^2\ge0\)
\(\Rightarrow P\ge0\)
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