Turning $2\le x$ into $\sqrt{1+\frac{4}{x^6}}\le \sqrt{5}$? I am supposed to turn $2\le x$ into $\sqrt{1+\frac{4}{x^6}}\le \sqrt{5}$, and I have no idea on how to approach this. I'll post my steps, even though I don't think they'll be of much help.
$$2\le x \rightarrow \frac{1}{x} \ge \frac{1}{2} \rightarrow\frac{4}{x}\ge2$$
and that is pretty much all I got. Thanks for all the help in advance!

In fact, we only need that $x \geq 1$:
\begin{align*}
x \geq 2
&\implies x \geq 1 \\
&\implies x^6 \geq 1^6 \\
&\implies \frac{1}{x^6} \leq \frac{1}{1^6} \\
&\implies \frac{4}{x^6} \leq \frac{4}{1^6} \\
&\implies 1 + \frac{4}{x^6} \leq 1 + \frac{4}{1^6} \\
&\implies \sqrt{1 + \frac{4}{x^6}} \leq \sqrt{1 + \frac{4}{1^6}} = \sqrt{5} \\
\end{align*}
