\(9a^2+b^2-6a+2b+5\)
\(=\left[\left(3a\right)^2-2.3.a+1\right]+\left(b^2+2b+1\right)+3\)
\(=\left(3a-1\right)^2+\left(b+1\right)^2+3\)
Ta thấy: \(\left(3a-1\right)^2\ge0;\left(b+1\right)^2\ge0\)\(\forall a;b\)
\(\Rightarrow\left(3a-1\right)^2+\left(b+1\right)^2+3>0\forall a;b\)
\(\Rightarrow9a^2+b^2-6a+2b+5>0\forall a;b\)