Có: \(x^2+x+1=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\)
Vì: \(\left(x+\frac{1}{2}\right)^2\ge0\)
=> \(\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\)
=> \(\frac{1}{\left(x+\frac{1}{2}\right)^2+\frac{3}{4}}\le\frac{4}{3}\)
=> \(-\frac{1}{x^2+x+1}\ge-\frac{4}{3}\)
=> \(1-\frac{1}{x^2+x+1}\ge-\frac{1}{3}\)
Vậy GTNN của bt đã cho là \(-\frac{1}{3}\) khi \(x=-\frac{1}{2}\)