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.

The task is:

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: $$\left(1+\frac{1}{n}\right)^n=1+n\cdot\frac{1}{n}+\frac{1}{2!}\cdot\frac{n(n-1)}{n^2}+...+\frac{1}{n!}\frac{n(n-1)...(n-(n-1))}{n^n}=$$ $$=2+\frac{1}{2!}\left(1-\frac{1}{n}\right)+...+\frac{1}{n!}\left(1-\frac{1}{n}\right)\left(1-\frac{2}{n}\right)...\left(1-\frac{n-1}{n}\right)<$$ $$<2+\frac{1}{2!}+...+\frac{1}{n!}<2+\frac{1}{2!}+...+\frac{1}{n!}+...=e.$$

  • $\begingroup$ 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.. $\endgroup$ – NotSoSmart97 Apr 20 at 18:01
  • $\begingroup$ @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. $\endgroup$ – Michael Rozenberg Apr 20 at 20:29
  • $\begingroup$ Wow, oh my gosh! You are a lifesaver! Thank you so much for the effort!! $\endgroup$ – NotSoSmart97 Apr 20 at 20:37
  • $\begingroup$ You are welcome! $\endgroup$ – Michael Rozenberg Apr 20 at 20:56

Hint: Simplify your inequality to $$\left(\frac{K+1}{K}\right)^{K+1}>\left(\frac{N-K}{N-K-1}\right)^{N-K-1}$$

  • $\begingroup$ Thanks for the fast reply! I tried that one. You basically can cancel out the N in the denominator and then divide over the (N-K-1)^{N-K-1). And since (K/N) is always lesser than 1 you can make that estimation. But how can I prove the remaining inequality now? $\endgroup$ – NotSoSmart97 Apr 20 at 16:59

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