On simplifying an inequality

Question

Me and my friend have been stuck on this problem for some time now. The problem is as follows:

Show that $$\Delta \cdot 2^{2\Delta} \geq 4n$$ $$ \implies \Delta \geq \frac12\log n - \frac12\log \log n + \frac12$$

Any help is appreciated. Thank you.

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Edit: Perhaps it'll be right to give some context to this question. Me and my friend are currently studying Chung et.al's paper On induced subgraph of the cube. The aforementioned problem of ours comes at the very last step of their paper.

Here $\Delta$ denotes the maximum degree of the graph $G$, an induced subgraph of the $n$-dimensional hypercube graph $Q^n$.

Answer 1

Suppose $n\ge 10^{0.1}$. Then $x>1$ and $\log(\log n) \ge -1$. Therefore

$$25^x>25x$$ $$10^{2x}>100(\frac{x}{4}2^{2x})\ge 100n$$ $$2x>\log n+2$$ $$2x>\log n-\log(\log n)+1.$$

Answer 2

If there are no restrictions on $n$ other than $n\ge 1$, then the result is false.

Let $x=2$ and $n$ tend to 1 from above.