\(y=\frac{2x+1}{x^2+2}\)
\(\Rightarrow y+\frac{1}{2}=\frac{2x+1}{x^2+2}+\frac{1}{2}\)
\(=\frac{2\left(2x+1\right)+x^2+2}{2\left(x^2+2\right)}\)
\(=\frac{4x+2+x^2+2}{2\left(x^2+2\right)}\)
\(=\frac{\left(x+2\right)^2}{2\left(x^2+2\right)}\)
Vì \(\left(x+2\right)^2\ge0\) với mọi x
\(2\left(x^2+2\right)\ge0\) với mọi x
\(\Rightarrow y+\frac{1}{2}\ge0\)
\(\Rightarrow y\ge-\frac{1}{2}\)
Dấu "=" xảy ra \(\Leftrightarrow x+2=0\)
\(\Leftrightarrow x=-2\)
Vậy GTNN của \(y=-\frac{1}{2}\) tại \(x=-2\)