Ta có: \(A=\frac{x^2-x+1}{x^2+x+1}\)
\(\Leftrightarrow Ax^2+Ax+A=x^2-x+1\)
\(\Leftrightarrow\left(A-1\right)x^2+\left(A+1\right)x+A-1=0\)
Để tồn tại \(x\Leftrightarrow\Delta\ge0\)
\(\Leftrightarrow\left(A+1\right)^2-4\left(A-1\right)\left(A-1\right)\ge0\)
\(\Leftrightarrow\left(3-A\right)\left(3A-1\right)\ge0\)
\(\Leftrightarrow\frac{1}{3}\le A\le3\)
Dấu " = " xảy ra \(\Leftrightarrow x=\frac{-\left(A+1\right)}{2\left(A-1\right)}=-1\)
Vậy \(A_{Min}=3\Leftrightarrow x=-1\)