I want to prove this statement:
$\alpha \geq1$ and $n\geq 2\alpha \log(2\alpha)$ then $n \geq \alpha \log(2n)$
It seemed pretty simple but I couldn't do too much. My first attempt was
$$ \log(n) \geq \log ( 2\alpha \log(2\alpha) ) $$
But this attempt didn't look promising. Could you give me a hint?
If ${a}\geq 1$, then we know that $\log(a)\geq0$ $\Rightarrow n\geq 0$. Substituting $ p = \alpha\log(2\alpha)$. We can clearly see the following: $$n\geq 2p$$ And because the numbers are positive then $$n\geq p$$ EDIT: Considering the change of what we have to prove: Here is the updated variant. If $n\geq 2\alpha\log(2\alpha)$, and all of the terms are positive, we can say that $n = 2\alpha\log(2\alpha) + \psi_1$ Where $\psi_1$ is some number greater than $0$. Plugging this into the second inequality we get what we have to prove: $$2\alpha\log(2\alpha) + \psi_1 = \alpha\log(4\alpha\log(2\alpha) + 2\psi_1) + \psi_2 $$ $${(2\alpha)}^{2\alpha}e^{\psi_1}=(4\alpha \log(2\alpha)+2\psi_1)^{a}e^{\psi_2}$$ $$(2\alpha)^{2\alpha}=((4\alpha\log(2\alpha)+2\psi_1)^{\frac{1}{2}})^{2a}e^{\psi_2-\psi_1}$$ $$e^{\frac{\psi_1-\psi_2}{2a}}=\left(\frac{\sqrt{(4\alpha\log(2\alpha)+2\psi_1)}}{2\alpha}\right) $$ Let $\psi_3=e^{\frac{\psi_1-\psi_2}{2a}}$, and we know that $\psi_3 \gt 0$, so we only have to prove: $$\left(\frac{\sqrt{(4\alpha\log(2\alpha)+2\psi_1)}}{2\alpha}\right) \gt 0$$ And since all of the terms are positive (including $\psi_1$), this will never equal or be less than $0$.
Assume, for the sake of contradiction, that there exists $(\alpha, n)$ such that $\alpha \ge 1$ and $n \ge 2\alpha \ln (2\alpha)$ and $n < \alpha \ln (2n)$.
We have $\alpha \ln(2n) > n \ge 2\alpha \ln(2\alpha)$ which results in $n \ge 2\alpha^2$. Thus, we have $n < \alpha \ln(2n) \le \sqrt{n/2}\ln(2n)$ which results in $\sqrt{2n} < \ln(2n)$, or $\mathrm{e}^{\sqrt{2n}} < 2n$. However, using $\mathrm{e}^x \ge 1 + x + \frac12 x^2 + \frac16 x^3$ for all $x > 0$, we have $$\mathrm{e}^{\sqrt{2n}} \ge 1 + \sqrt{2n} + \frac12 \sqrt{2n}^2 + \frac16 \sqrt{2n}^3 > 2n.$$ Contradiction.
