\(\dfrac{x^2+x+1}{x^2+2x+1}=\dfrac{\left(x+1\right)^2-\left(x+1\right)+1}{\left(x+1\right)^2}=1-\dfrac{1}{x+1}+\dfrac{1}{\left(x+1\right)^2}\)
Đặt y=1/x+1=>\(1-y+y^2=\left(y-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\)
Vậy GTNN của \(\dfrac{x^2+x+1}{x^2+2x+1}\) là \(\dfrac{3}{4}\) tại x=1