-1
$\begingroup$

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

$\endgroup$
1
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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
0
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.