Question

Chứng minh a\(^2\)+b\(^2\)+3> ab+a+b với mọi a,b

Giúp mình nhé

Answer 1

\(a^2+b^2+3>ab+a+b\)

\(\Leftrightarrow2\left(a^2+b^2+3\right)>2\left(ab+a+b\right)\)

\(\Leftrightarrow\left(a^2-2a+1\right)+\left(b^2-2b+1\right)+\left(a^2-2ab+b^2\right)+4>0\)

\(\Leftrightarrow\left(a-1\right)^2+\left(b-1\right)^2+\left(a-b\right)^2+4>0\) \(\forall a,b\)

Vậy \(a^2+b^2+3>ab+a+b\forall a,b\)