How to prove $n^{3/2} \le 2^n/n$ for all $n\geq 8$ via Induction?

Question

How to prove $n^{3/2} \le 2^n/n$ for all $n \geq 8$

My attempt:

$\textbf{Base case}$ is $n_0=8$: $\quad \left(8^{3/2}=4\right)\leq \left(32=\frac{2^8}{8}\right) \checkmark$

$\textbf{Induction hypothesis}$: $\exists n\in \mathbb{N}_{\geq 8}:(n+1)^{3/2}\leq \frac{2^{n+1}}{n+1}$

$\textbf{Inductive step}$: $$ \begin{gather} (n+1)^{3/2}\leq \frac{2^{n+1}}{n+1} \end{gather} \\ \iff (n+1)^{5/3}\leq 2^{n+1} \\ \iff (n+1)^{5/3}\leq 2^n \cdot 2 \quad |\uparrow ^{3/5} \\ \iff n+1\leq 2^{3n/5}\cdot 2^{3/5}\ $$ This, sadly, doesn't get me anywhere, any ideas?

Answer 1

Better is to prove $$n^{5/2}\le 2^n$$, then we have to prove $$(n+1)^{5/2}\le 2^{n+1}$$. Then we get by multiplying by $2$: $$2^{n+1}\geq 2\cdot n^{5/2}$$ and it remaines to show that $$2\cdot n^{5/2}\geq (n+1)^{5/2}$$

Answer 2

Consider function $f(x)=4^x-x^5$. Now we have to prove that $f(x)>0$ for $x\geq8$ using induction.

Let start with $f(8)=4^8-8^5=2^{16}-2^{15}\gt0$

Now, for any $x=k\geq8$ let us assume $f(k)=4^k-k^5\geq0$

Now put $x=k+1$, then $f(k+1)=4^{k+1}-(k+1)^5=4\times4^k-(k+1)^5$

$f(k+1)=[4^k-k^5]+[4^k-5k^4]+[4^k-10k^3]+[4^k-10k^2-5k^-1]$

$f(k+1)=[4^k-k^5]+[4^k-\frac{5}{k}k^5]+[4^k-\frac{10}{k^2}k^5]+[4^k-(\frac{10}{k^3}+\frac{5}{k^4}+\frac{1}{k^5})k^5]$

$f(k+1)=t_1+t_2+t_3+t_4$ (Assume all one square bracket as a single term)

since we are assuming $t_1\geq0$ and in other terms, the quotient of $k^5$ in $t_2,t_3,t_4$ is $\lt1$ which makes positive part more dominant.

Hence, all the terms are positive giving $f(k+1)\geq0$

Now, by theorem of Induction, $f(k)\geq0$ for $x\geq8$.

Hence Proved!

Answer 3

Continuing Sonnhard's answer. Hint: $$ (2n^{5/2})^{2} = 4 n^{5} \ge n^{5} + n^{5} + n^{5} + n^{5} \ge n^{5} + 8 n^{4} + 64 n^{3} + 8^{3} n^{2} $$ Or can also $$n^{5} + n^{5} + n^{5} + n^{5} \ge n^{5} + 16 n^{4} + 64 n^{3} $$

---

$$ (n+1)^{5} = n^{5} + \binom{5}{1}n^{4} + \binom{5}{2} n^{3} + \binom{5}{4} n^{2} +1$$