How can I show that $O( \sqrt{n} \cdot \log_2(n) \cdot \log_2(\log_2(n))) < O(n)$?

Question

The question is pretty straightforward: How can I show that

$$O( \sqrt{n} \cdot \log_2(n) \cdot \log_2(\log_2(n))) < O(n)$$

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The question can be reduced by observing that $O(n) = O(\sqrt{n}) \cdot O(\sqrt{n})$.

Then we want to show that

$$O( \sqrt{n} \cdot \log_2(n) \cdot \log_2(\log_2(n))) < O(\sqrt{n}) \cdot O(\sqrt{n})$$

$$ \require{cancel} \cancel{O( \sqrt{n})} \cdot O(\log_2(n) \cdot \log_2(\log_2(n))) < O(\sqrt{n}) \cdot \cancel{O(\sqrt{n})}$$

So what is left is to prove that $$ O(\log_2(n) \cdot \log_2(\log_2(n))) < O(\sqrt{n})$$

But since I cannot find a relation between $\log_2(n)$ and $\sqrt{n}$, I couldn't prove it yet.

Answer 1

Show that $\lim_{n \rightarrow \infty} \frac{n}{\sqrt{n} (\log n) (\log \log n)} = \infty$, by applying L'Hospital's rule twice.