I saw many proofs that the sequence $(1+\frac{1}{k})^{k+\frac{1}{2}}$ for $k \in \mathbb{N}$ is strictly decreasing but all of them uses derivatives or integrals and I wonder if there is a purely algebraic proof (without derivatives or any theorem that uses derivative as a part of their proof, like integral).
Here is my attempts:
We need to prove that:
$$a_{k-1}>a_{k} $$
$${a_{k-1}}^2>a_{k}^2$$
$$\left( \frac{k}{k-1} \right)^{2k-1}> \left( \frac{k+1}{k} \right)^{2k+1}$$
$$(k)^{4k} >({k+1})^{2k+1} ({k-1})^{2k-1}$$
Here I tried to use the binomial theorem to prove that $(k)^{4k} >({k+1})^{2k+1} ({k-1})^{2k-1}$, or in other form $\left(1+ \frac{1}{k^2-1} \right)^{2k} > 1+\frac{2}{k-1}$, but the sequence $\left(1+ \frac{1}{k^2-1} \right)^{2k} - 1-\frac{2}{k-1}$ converge TOO fast to $0 $ and after a lot of calculations I realised that I need to use most or all the terms in the expatiation of $\left(1+ \frac{1}{k^2-1} \right)^{2k}$ so that the RHS >LHS, so the this approach lead to a dead end.
I also tried to use induction, but I couldn't get to work, It seems it is too hard or impossible to use induction in this sequence ,
I noticed that the inequality $a_{k-1}>a_{k} $ implies that $ \frac{k}{\sqrt{(k+1)(k-1)}}>\left( \frac{k+1}{k-1}\right)^{\frac{1}{4k}}$ which is an inequality include $\frac{AM}{GM}$. (I don’t know if this is useful, but it was very interesting and for me it was a dead end.)
I also tried to to prove $ \frac{2k^3}{{2k (k+1)(k-1)} }>\left( \frac{k+1}{k-1}\right)^{\frac{1}{2k}}$ (I multiplied the RHS bt $\frac{2k}{2k}$ to use the AM-GM inequality) the RHS will be $$\frac{2k +\frac{1}{k-1} +\frac{1}{k+1} }{2k}$$ I tried to find an $2k$ sequence of positive numbers st their sum is ${2k +\frac{1}{k-1} +\frac{1}{k+1} }$ and their product is $ \frac{k+1}{k-1}$ to use the AM-GM inequality but I also couldn't.
Unfortunately, all my efforts to prove this inequality were unsuccessful and I couldn’t make any progress at all, so I decided to ask here for help.
