Lời giải:
Ta có:
\((x+1)^2\left (\frac{1}{x^2}+\frac{2}{x}+1\right)=(x+1)^2\left(\frac{1}{x}+1\right)^2\)
\(=\frac{(x+1)^4}{x^2}\)
Áp dụng BĐT AM-GM:
\(x+1\geq 2\sqrt{x}\Rightarrow (x+1)^4\geq 16x^2\)
Do đó \((x+1)^2\left (\frac{1}{x^2}+\frac{2}{x}+1\right)=\frac{(x+1)^4}{x^2}\geq \frac{16x^2}{x^2}=16\)
Dấu bằng xảy ra khi \(x=1\)