# Prove that if $\alpha \geq1$ and $n\geq 2\alpha \log(2\alpha)$ then $n \geq \alpha \log(2n)$

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?

• I don't see $a$ anywhere in the final result or in the other condition. So how are we expected to use $a \ge 1$? Commented Mar 10 at 9:16
• @Haris Yeah, it was a typo ($\alpha$ ) Commented Mar 10 at 9:19

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 $$0$$.

• I´m sorry. I made a mistake. I have to prove that $n \geq \alpha log(2n)$ Commented Mar 10 at 9:21
• @Masd I have a question. I think that $\left(\frac{\sqrt{(4\alpha\log(2\alpha)+2\psi_1)}}{2\alpha}\right) > 0$ and $e^{\frac{\psi_1-\psi_2}{2a}} > 0$ does not imply $\psi_2 \ge 0$. You need to prove that $\psi_2 < 0$ and $e^{\frac{\psi_1-\psi_2}{2a}}=\left(\frac{\sqrt{(4\alpha\log(2\alpha)+2\psi_1)}}{2\alpha}\right) > 0$ is impossible. In other words, you need to prove that $$e^{\frac{\psi_1}{2a}}\ge \left(\frac{\sqrt{(4\alpha\log(2\alpha)+2\psi_1)}}{2\alpha}\right).$$ Commented Mar 11 at 0:37
• By definition $\psi_2$ is bigger than 0
– Masd
Commented Mar 11 at 13:54
• @Masd I think that you have to prove that there always exists $\psi_2\ge 0$ such that $2\alpha\log(2\alpha) + \psi_1 = \alpha\log(4\alpha\log(2\alpha) + 2\psi_1) + \psi_2$; in other words, $2\alpha\log(2\alpha) + \psi_1 = \alpha\log(4\alpha\log(2\alpha) + 2\psi_1) + \psi_2$ implies $\psi_2 \ge 0$ (this is what you want to prove rather than definition). Commented Mar 11 at 23:24
• I was just rephrasing the statement that what we have to prove, is all
– Masd
Commented Mar 11 at 23:36

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.

• I don't understand the last inequality. Is it due to $(\frac{1}{2})\sqrt{2n}^{2} = n$? Commented Mar 11 at 5:23
• @xenuti By AM-GM, we have $\sqrt{2n} + \frac16 \sqrt{2n}^3 \ge \sqrt{2n} + \frac{1}{16} \sqrt{2n}^3 \ge 2\sqrt{ \sqrt{2n} \cdot \frac{1}{16} \sqrt{2n}^3 } = n$. Also, $\frac12 \sqrt{2n}^2 = n$. Commented Mar 11 at 6:13