How to prove that for any $n$ in $\mathbb{N}$ that $(\frac{3}{2})^n \ge n$?

Question

Well, I was trying to do that using proof by induction and my attempt is :
Base case : $(\frac{3}{2})^0 \ge 0$, true
Assumption : $(\frac{3}{2})^k \ge k$.
I've multiplied both sides by $(\frac{3}{2})$ i got $(\frac{3}{2})^{k+1}\ge (\frac{3}{2})k$ i don't know what do to tun this $(\frac{3}{2})k$ to $k+1$

By the way : (for any $n\in\mathbb{N}$)

Answer 1

Once you have $\left(1+\frac{1}{2}\right)^k > k$ for $k\geq 2$, you get that $$\left(1+\frac{1}{2}\right)^{k+1}=\left(1+\frac{1}{2}\right)\left(1+\frac{1}{2}\right)^k > \left(1+\frac{1}{2}\right)k = k+\frac{k}{2} \geq k+1.$$

Answer 2

$\frac{3}{2} \cdot k \geq k +1 \Leftrightarrow 3k \geq 2k +2 \Leftrightarrow k \geq 2$

Answer 3

Here is another way:

Define $f(x)=\left(\frac{3}{2}\right)^x-x$. We need to show that $f(n)\ge 0$ for all $n\in \mathbb{N}$. By inspection, it is true for $n=0, 1, 2, 3$. Since $$f'(x)= \left( \frac{3}{2}\right)^x\ln \frac{3}{2}-1 \ge 0, \quad \forall x\ge 3,$$ the function is increasing on $[3,\infty)$. As $f(3)\ge 0$, the result follows.