TITLE: How can I prove this monster inequality?
QUESTION [7 upvotes]: Given fixed values $\alpha \in (0,1)$ and $ n \in \Bbb N$.
For $k \in \{1,...,n-1\}$, consider the function $$f(k)=\left(1-\alpha\sqrt{\frac{n-k}{k(n-1)}}\right)^k\left(1+\alpha\sqrt{\frac{k}{(n-k)(n-1)}}\right)^{n-k}$$.
How can I show that, for all $k \in \{1,...,n-1\}$, $$f(k) \ge \left(1-\alpha\right)\left(1+\frac{\alpha}{n-1}\right)^{n-1}$$ (where equality occurs exactly at $k=1$)?
EDIT 1: I tried showing that $f$ is monotoniacally increasing in $k$: $$\frac{d}{dk} \log (f(k))= \log\left(1-\alpha\sqrt{\frac{n-k}{k(n-1)}}\right)-\log\left(1+\alpha\sqrt{\frac{k}{(n-k)(n-1)}}\right)+\frac{\alpha n}{2(\sqrt{k(n-k)(n-1)}-\alpha (n-k))}+\frac{\alpha n}{2(\sqrt{k(n-k)(n-1)}+\alpha k)}$$ which should be $\ge 0$, but I don't see how to make the right estimates.
EDIT 2: To the background of this inequality: This is what remains to be shown when you apply Lagrange multipliers to this inequality. If anyone has an idea how to procede in that inequality without using Lagrange multipliers, I would be equally content.

REPLY [5 votes]: Edit: I found a new method to solve it.
$F=\log\left(1-\alpha\sqrt{\frac{n-k}{k(n-1)}}\right)-\log\left(1+\alpha\sqrt{\frac{k}{(n-k)(n-1)}}\right)=\log\left(\dfrac{\sqrt{k(n-k)(n-1)}-\alpha (n-k)}{\sqrt{k(n-k)(n-1)}+\alpha k} \right)$
let $x=\dfrac{2(\sqrt{k(n-k)(n-1)}-\alpha (n-k))}{\alpha n}, x+2=\dfrac{2(\sqrt{k(n-k)(n-1)}+\alpha k)}{\alpha n}$
$F=\log\left(\dfrac{x}{x+2}\right), h(x)=\log\left(\dfrac{x}{x+2}\right)+\dfrac{1}{x}+\dfrac{1}{x+2}$
if we prove $h(x)>0$,then the problem solved.
$h'(x)=\dfrac{1}{x}-\dfrac{1}{x+2}-\dfrac{1}{x^2}-\dfrac{1}{(x+2)^2}=-\dfrac{4}{x^2(x+2)^2} <0$
$h(x) $is mono decreasing, $h(x)_{min}=h(+\infty)=0 \to h(x)>0$
thus $\dfrac{d}{dk} \log (f(k))>0$