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For $n \geq 2$, we want to prove that

$$ \prod_{k=0}^n {n \choose k} \leq \left( \dfrac{ 2^n - 2 }{n-1} \right)^{n-1} $$

My first approach was to try and use induction on $n$. The base case is trivial $2 \leq 2$ and all boils down to prove that

$$ \left( \dfrac{ 2^n - 2 }{n-1} \right)^{n-1} { n+1 \choose n} \leq \left( \dfrac{ 2^{n+1} - 2 }{n} \right)^{n} $$

Which seems a bit cumbersome. Then I realize we can do the following. We have $n-1$ products of non 1's in both sides of the equation and thus if we can prove that

$$ {n \choose k} \leq \dfrac{ 2^n - 2}{n-1} $$

for $k = 1,...,n-1$ then we are basically done. This indeed seems as a more handeable inequality than the one arising from the induction hypothesis. Notice that ${n \choose k}$ is maximal when $k = [n/2]$ (floor function). But I dont see a way to do it analytically. But, we can also do this:

Rewrite it as

$$ \dfrac{ (n-1) {n \choose k } }{2^n - 2 } \leq 1 $$

Can we interpret the LHS as a probability? We see that the denominator is the size of same sample space. Since $2^n = \sum {n \choose k}$ counts all possible commitees then $2^n - 2$ is the size of the sample space of all possible committees of $n$ people where we donde consider the empty commitee and the commitee with all $n$ people (so that takes 2 out of $2^n$). Now the numerator might be the size of some event of sample space. Event: commitees of size $k$ out of $n$ people. Then the LHS is a $P(E)$ and we know $P(E) \leq 1$. An so this proves what we wanted. But, I am not 100 % whether this argument might work. Any guidance would be helpful!

James
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1 Answers1

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As failedentertainment noted, your approach cannot work, since your intermediate goal is false. Note that $\prod_{k=0}^n\binom{n}k=\prod_{k=1}^n\binom{n}k$, so you want to prove that

$$\prod_{k=1}^{n-1}\binom{n}k\leq\left(\dfrac{2^n-2}{n-1}\right)^{n-1}\;,\tag{1}$$

where each side is a product of $n-1$ factors. Take $(n-1)$-st roots on both sides: $(1)$ is equivalent to

$$\left(\prod_{k=1}^{n-1}\binom{n}k\right)^{\frac1{n-1}}\le\frac{2^n-2}{n-1}\;.$$

Now use the arithmetic mean-geometric mean inequality. (You may have to think a little to see why it applies.)

Brian M. Scott
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  • Because $2^n = \sum {n \choose k }$ ? – James May 03 '20 at 04:04
  • @James: Yes, and taking $k$ from $1$ to $n-1$ cuts out two of the $2^n$ subsets. – Brian M. Scott May 03 '20 at 04:06
  • Prof. you have a typo: should. be $ \prod ^{1/(n-1) } $ – James May 03 '20 at 04:07
  • @James: Good catch! You’re absolutely right; thanks! – Brian M. Scott May 03 '20 at 04:08
  • Yes Prof: $\sum_{k=1}^{n-1} = \sum_{k=0}^n - 2 = 2^n - 2 $ – James May 03 '20 at 04:09
  • Very nice proof! My proof sucks... Was I on the right track? I would like some feedback on my attempt.. or it was completely bad? – James May 03 '20 at 04:10
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    @James: You got off track very early with $$\left(\frac{2^n-2}{n-1}\right)^{n-1}\binom{n+1}n\le\left(\frac{2^{n+1}-2}n\right)^n;.$$ It looks like you were forgetting that when you go from $\prod_{k=0}^n\binom{n}k$ to $\prod_{k=0}^{n+1}\binom{n+1}k$, you’re not just multiplying by $\binom{n+1}n$: every binomial coefficient in the product now has $n+1$ as its upper number instead of $n$. – Brian M. Scott May 03 '20 at 04:16