\(A=\frac{x+1}{x^2}=\frac{1}{x}+\frac{1}{x^2}\)
\(=\frac{1}{x^2}+2\cdot\frac{1}{x}\cdot\frac{1}{2}+\frac{1}{4}-\frac{1}{4}\)
\(=\left(\frac{1}{x}+\frac{1}{2}\right)^2-\frac{1}{4}\ge-\frac{1}{4}\forall x\)
Dâu "=" \(\Leftrightarrow\left(\frac{1}{x}+\frac{1}{2}\right)^2=0\Leftrightarrow x=-2\)