<s>[INST] How to see this inequality Consider
$$
\frac{dx}{dt}=\sqrt{x^2+1}+t^2,\qquad x\in\mathbb{R}.
$$
Then I do not see why it is
$$
\lvert\frac{dx}{dt}\rvert\leq 2\lvert x\rvert+t^2.
$$
 [/INST] The steps that come to mind are
$$\begin{array}{}
\frac{dx}{dt}=\sqrt{x^2+1}+t^2 \\
|\frac{dx}{dt}|=|\sqrt{x^2+1}+t^2| \\
|\frac{dx}{dt}|\le|\sqrt{x^2+1}|+|t^2| & \qquad \text{Triangle Inequality} \\
|\frac{dx}{dt}|\le \sqrt{x^2+1} + t^2  \\
\end{array}$$
If we had $\sqrt{x^2+1} \le 2|x|$, then we could get your final result but
$$\begin{array}{}
\sqrt{x^2+1} \le 2|x| \\
x^2+1 \le 4x^2 \\
3x^2 \ge 1 \\
x^2 \ge \frac{1}{3} \\
|x| \ge \frac{1}{\sqrt3}
\end{array}$$
So we can say that
$$|\frac{dx}{dt}|\le \ 2|x| + t^2, \qquad \quad |x| \ge \frac{1}{\sqrt3}$$
</s>