Đáp án: $m \ge - \sqrt 2 $
Giải thích các bước giải:
$\begin{array}{l}
m\sqrt {2{x^2} + 1} + 2x \ge 0\forall x\\
\Rightarrow \frac{{m\sqrt {2{x^2} + 1} }}{x} + 2 \ge 0\forall x\\
\Rightarrow m\sqrt {\frac{{2{x^2} + 1}}{{{x^2}}}} + 2 \ge 0\forall x\\
\Rightarrow m \ge - \frac{2}{{\sqrt {2 + \frac{1}{{{x^2}}}} }}\forall x\\
\Rightarrow m \ge \max - \frac{2}{{\sqrt {2 + \frac{1}{{{x^2}}}} }}\forall x\\
Do:{x^2} \ge 0\forall x\\
\Rightarrow \sqrt {2 + \frac{1}{{{x^2}}}} \le \sqrt 2 \\
\Rightarrow \frac{2}{{\sqrt {2 + \frac{1}{{{x^2}}}} }} \ge \frac{2}{{\sqrt 2 }} = \sqrt 2 \\
\Rightarrow - \frac{2}{{\sqrt {2 + \frac{1}{{{x^2}}}} }} \le - \sqrt 2 \\
\Rightarrow max - \frac{2}{{\sqrt {2 + \frac{1}{{{x^2}}}} }} = - \sqrt 2 \\
\Rightarrow m \ge - \sqrt 2
\end{array}$
