Prove $ \forall c \in \mathbb{R^{+}}, \forall B \in \mathbb{N}, \exists n \in \mathbb{N}, (n \geq B) \wedge (\sqrt{2n} > c \ln(n+2)) $?

Question

So far I've proved the following: $2n \geq \ln(2n) > \ln(n+2)$ for $n \geq 3 $

The proof must be without the use of calculus but I have no idea on how to proceed.

Answer 1

We have: $\sqrt{2n} > c\ln(n+2)\iff \sqrt{2n} > c\ln2 + c\ln(1+n/2)$ . Put $ m = n/2 \implies 2\sqrt{m}> c\ln 2+ c\ln(1+m)\iff \dfrac{2\sqrt{m}}{c}> \ln2 +\ln(1+m) \iff d\sqrt{m} > \ln2 + \ln(1+m), d = \dfrac{2}{c}> 0$. Observe $\ln(1+m) < \dfrac{d\sqrt{m}}{2}\iff 1+m < e^{\frac{d\sqrt{m}}{2}}$. Now put $ t = \dfrac{d\sqrt{m}}{2}\implies m = \dfrac{4t^2}{d^2}\implies 1+\dfrac{4t^2}{d^2} < e^t\implies 1+c^2t^2 < e^t$. But $e^t > 1+t+t^2/2+t^3/6> 1+t^3/6, t > 0\implies t^3/6 > ct^2\implies t > 6c\implies \dfrac{d\sqrt{m}}{2} > 6c\implies \dfrac{\sqrt{m}}{c} > 6c\implies \sqrt{m} > 6c^2\implies m > 36c^4\implies n > 72c^4$. Thus let $B_1 = 72c^4\implies d\sqrt{m}> \ln 2+\dfrac{d\sqrt{m}}{2}\implies d\sqrt{m} > \ln4\implies m > \dfrac{(\ln 4)^2}{d^2}= \dfrac{c^2(\ln 4)^2}{4} \implies n > \dfrac{c^2(\ln 4)^2}{2} = B_2$. Thus we can take $B = \text{max}(B_1,B_2)= \text{max}\left(72c^4, \dfrac{c^2(\ln 4)^2}{2}\right)$, and this yields the sought inequality.

Note: My answer uses the unedited post, but it can still be modified slightly to accommodate the edited post.

Answer 2

Let us recall the well-known fact that an exponential function grows faster than any polynomial function. To be specific, we just need the fact that
$$ \forall c_0\in\Bbb{R^+},\,\exists m_0\in\Bbb{N},\,\forall x\in \mathbb{R},\,(x\gt m_0)\Rightarrow (e^x > x^{c_0})$$

Let $c_0 =2c$. Then there exists $m_1\in\Bbb N$, $\forall x\in \mathbb{R},\,\,x\gt m_1\Rightarrow (e^x > x^{2c})$. So, $\forall y\in\Bbb R,$ $$\begin{align}&y\gt {m_1}^2/2\\ \Rightarrow\ &\sqrt{2y} \gt m_1\\ \Rightarrow\ &e^{\sqrt{2y}} > {\sqrt{2y}}^{2c} =(2y)^c\\ \Rightarrow\ &\sqrt{2y} > c\ln(2y)\\ \end{align}$$

So, if we choose sufficiently large $n\in\Bbb N$ such that $n\gt 2$ and $n \gt B$ and $n \gt {m_1}^2/2$, we will have $ \sqrt{2n} > c\ln(2n)) > c\ln(n+2) $.