8
$\begingroup$

This question already has an answer here:

I proved this inequality in the following way:

Lemma: $r \in \Bbb N, r \geq 3$. We have $r^r \gt (r+1)^{r-1}$.
Proof: We apply the AM-GM inequality to the $r$ positive integers where there are $r-1$ $(r+1)$'s and one $1$. We obtain $$\frac {1+(r-1)(r+1)}{r} \gt ((r+1)^{r-1})^{\frac {1}{r}}$$ wherefrom we get (since the $r^{{\rm th}}$ power function is increasing), $r^{r} \gt (r+1)^{r-1}$.

Now, I used mathematical induction to prove the statement. We have from the lemma, $(k!)^{2} \gt k^k \implies (k!)^{2} \gt (k+1)^{k-1}$ and multiplying this inequality by $(k+1)^{2}$, we have $((k+1)!)^{2} \gt (k+1)^{k+1}$ and obviously $(3!)^{2} \gt 3^3$.

Is there any direct proof of the statement that does not use induction and calculus?

$\endgroup$

marked as duplicate by Martin Sleziak, JonMark Perry, choco_addicted, user91500, M. Vinay Jun 4 '16 at 8:54

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ See also Stirling's approximation. $\endgroup$ – Lucian May 8 '14 at 0:24
  • $\begingroup$ Well, Stirling's approximation with the upper bound on the error from the quadrature formula to ensure it starts working at n=3. $\endgroup$ – Adam Hughes Jun 28 '14 at 9:28
  • 4
    $\begingroup$ See also this question and other posts linked there $\endgroup$ – Martin Sleziak Jun 4 '16 at 7:15
9
$\begingroup$

Hint : $$n \le r(n-r+1)$$

Now take the product for $r=1$ to $r=n$.

$\endgroup$
  • $\begingroup$ Ah... I see $(n-r)(r-1) \geq 0$. I should have seen this before. $\endgroup$ – Indrayudh Roy May 7 '14 at 20:45

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