Important: Proof of this inequality (?)

I'm really screwed since I don't know how to prove this inequality. I've tried creating an $$\epsilon$$ surrounding and everything but nothing seems to work.

Prove $$\left(\frac{K+1}{N}\right)^{K+1}\left(\frac{N-K-1}{N}\right)^{N-K-1}> \left(\frac{K}{N}\right)^{K+1}\left(\frac{N-K}{N}\right)^{N-K-1}$$ Where $$N >>K$$, also $$N>K+1$$ and $$N, K$$ are positive integers.

Does anyone have an idea? Thanks!!

PS: I'm not sure, if I'm allowed to, but if there is no other way, you may use: N+1=2K

It's $$\left(1+\frac{1}{k}\right)^{k+1}>\left(1+\frac{1}{n-k-1}\right)^{n-k-1},$$ which is true because $$\left(1+\frac{1}{k}\right)^{k+1}>e>\left(1+\frac{1}{n-k-1}\right)^{n-k-1}.$$ Because for all natural $$n$$ we obtain: $$\left(1+\frac{1}{n}\right)^n=1+n\cdot\frac{1}{n}+\frac{1}{2!}\cdot\frac{n(n-1)}{n^2}+...+\frac{1}{n!}\frac{n(n-1)...(n-(n-1))}{n^n}=$$ $$=2+\frac{1}{2!}\left(1-\frac{1}{n}\right)+...+\frac{1}{n!}\left(1-\frac{1}{n}\right)\left(1-\frac{2}{n}\right)...\left(1-\frac{n-1}{n}\right)<$$ $$<2+\frac{1}{2!}+...+\frac{1}{n!}<2+\frac{1}{2!}+...+\frac{1}{n!}+...=e.$$
• Ah, thanks! But $(1+\frac{1}{k})^{k+1}$ > e holds because $(1+\frac{1}{k})^{k}$ in the limit for k goes towards e, but does this also hold not in the limit? I'm sorry, but I don't get the second inequality about why e is bigger than the right term.. – NotSoSmart97 Apr 20 at 18:01
• @NotSoSmart97 For the left inequality prove that for all $x>0$ we have $\left(1+\frac{1}{x}\right)^{x+1}>e.$ The proof of the right inequality we can get by the same way or see my post. – Michael Rozenberg Apr 20 at 20:29
Hint: Simplify your inequality to $$\left(\frac{K+1}{K}\right)^{K+1}>\left(\frac{N-K}{N-K-1}\right)^{N-K-1}$$