A Notorious Inequality Problem-1

We will prove that it holds for all $n\in \mathbb{N},n-{1}$ . Consider the expression $(1+\frac{1}{1})(1+\frac{1}{2})...(1+\frac{1}{n})$ this expression is $n+1$ . Also, $(1+\frac{1}{1})(1+\frac{1}{2})...(1+\frac{1}{n})>(\frac{2}{\sqrt{1}})(\frac{2}{\sqrt{2}})...(\frac{2}{\sqrt{n}})=\frac{2^n}{\sqrt{n!}}$ so, we can conclude that $n+1>\frac{2^n}{\sqrt{n!}}\implies (n+1)\sqrt{n!}>2^n$

Let use Induction to show the above inequality holds true for all positive integers $n \ge 2$ . If we rewrite it as $n! > \frac{4^n}{(n+1)^2}$ , then we have:

CASE I ( $n= 2$ ): $2! > \frac{4^2}{(2+1)^2} \Rightarrow 2 > \frac{16}{9} \approx 1.78 \Rightarrow$ (TRUE).

CASE II $(n = k): k! > \frac{4^k}{(k+1)^2} \Rightarrow$ (ASSUMED TRUE).

CASE III $(n=k+1):$ $(k+1) \cdot k! > \frac{4^k}{(k+1)^2} \cdot (k+1);$

or $(k+1)! > \frac{4^k}{k+1} \cdot \frac{4(k+2)^2}{4(k+2)^2};$

or $(k+1)! > \frac{4^{k+1}}{[(k+1)+1]^2} \cdot \frac{(k+2)^2}{4(k+1)};$

or $(k+1)! > \frac{4^{k+1}}{[(k+1)+1]^2} \cdot \frac{k^2+4k+4}{4(k+1)};$

or $(k+1)! > \frac{4^{k+1}}{[(k+1)+1]^2} \cdot [1 + \frac{k^2}{4(k+1)}] > \frac{4^{k+1}}{[(k+1)+1]^2};$

or $\boxed{(k+1)! > \frac{4^{k+1}}{[(k+1)+1]^2}} \Rightarrow$ (TRUE).

$\mathbb{Q.} \mathbb{E.} \mathbb{D.}$