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It asks to prove the above statement, and I was given a hint to use the AM$\ge$GM inequality, i.e. the geometric mean is always less than or equal to the arithmetic mean: $(a_1a_2· · · a_n)^\frac{1}{n} ≤ \frac{a_1 + a_2· · · + a_n}{n}$, with equality if and only if $a_1=a_2=...=a_n$.

I tried to use induction to prove this but got stuck on proving the inductive case, if $P(n)$ holds then $P(n+1)$ holds.

Could anyone help me in trying to prove this or at least set me on the right path?

Thanks in advance!

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  • $\begingroup$ This clearly holds asymptotically; by convexity, if it holds for some $n$ it holds for all larger $n$ (for $n \ge 1$ at least) and the inequality holds for $n = 2$ so we are done. $\endgroup$ – Brevan Ellefsen Aug 10 '19 at 5:02
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By the AM/GM inequality applied to $1,2,\dots,n$ we have $$\sqrt[n]{n!}<\frac{1+2+\dots+n}n$$ This can be rearranged (using the triangular sum $1+2+\dots+n=\frac{n(n+1)}2$) to get $$n!<\left(\frac{n(n+1)}{2n}\right)^n$$ $$n!<\left(\frac{n+1}2\right)^n$$

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    $\begingroup$ Oh that is much simpler than what I was trying to write, thank you so much! $\endgroup$ – mathnerd Aug 10 '19 at 4:49
  • $\begingroup$ @Alex Now, please accept my answer. Click the green tick beside it. $\endgroup$ – Parcly Taxel Aug 10 '19 at 4:50
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We can simply apply the $AM -GM$ inequality for first n natural numbers as they are all positive

$GM ≤ AM$ Now $GM \to \sqrt[n]{\prod^n_{i=1} i} = \sqrt[n]{n!}$

$AM \to \frac{\sum_{r=1}^n r}{n} = \frac{\frac{n(n+1)}{2}}{n}= \frac{(n+1)}{2} $

Now $AM-GM$ gives $\sqrt[n]{n!} ≤ \frac{(n+1)}{2}$ But as $n≥ 2$ and as you point out in the hint of the question, the equality will never hold. So it is safe to write $$\sqrt[n]{n!} < \frac{(n+1)}{2}$$ or $$n! < ( \frac{(n+1)}{2})^n $$

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    $\begingroup$ Thank you so much!! $\endgroup$ – mathnerd Aug 10 '19 at 4:50
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You can also break this down to two-number inequalities, combining factors symmetrically from each end of the sequence $1,2,...,n-1,n$. Take the factorial twice to avoid to have to split in the middle, to get $$ (n!)^2=\prod_{k=1}^nk(n+1-k)\le\prod_{k=1}^n\left(\frac{n+1}2\right)^2, $$ the last step using $0\le(\sqrt a-\sqrt b)^2\implies \sqrt{ab}\le\frac{a+b}2$.

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