<s>[INST] 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!
 [/INST] Your first implication $2\leq x\implies \frac{1}{x}\geq\frac{1}{2}$ is actually false: you have to flip the inequality direction, roughly because taking the reciprocal of a large number will give you a small number. Here is how you can do it:
\begin{align*}
2 &\leq x \\
\implies \frac{1}{x} &\leq \frac{1}{2} \\
\implies \frac{1}{x^6} &\leq \frac{1}{64} \\
\implies \frac{4}{x^6} &\leq \frac{1}{16} \\
\implies 1+\frac{4}{x^6} &\leq \frac{17}{16}<5 \\
\implies \sqrt{1+\frac{4}{x^6}} &< \sqrt{5}
\end{align*}
where taking the square root accross the inequality was justified because everything is positive.
</s>