Ta có:
$\begin{array}{l} {\left( {x - 1} \right)^2} \ge 0,\forall x\\ \Leftrightarrow {x^2} - 2x + 1 \ge 0\\ \Leftrightarrow {x^2} + 1 \ge 2x\\ \Leftrightarrow {x^2} + 2x + 1 \le 2\left( {{x^2} + 1} \right)\\ \Leftrightarrow {\left( {x + 1} \right)^2} \le 2\left( {{x^2} + 1} \right)\\ \Rightarrow \frac{{2{x^2} + 2}}{{{{\left( {x + 1} \right)}^2}}} \ge \frac{{2{x^2} + 2}}{{2\left( {{x^2} + 1} \right)}}\\ \Leftrightarrow Q \ge 1 \end{array}$
Dấu '=' xảy ra khi :$ x - 1 = 0 \Rightarrow x = 1$
$Vậy {Q_{\min }} = 1$
