Proof writing for an inequality

Question

Prove that:

$$(n!)^2>n^n$$

Also prove that, $$2×4×6......(2n)<(n+1)^n$$

I haven't solved any inequalities of this kind before so all suggestions are welcome.

Answer 1

[Note: $(1!)^2 = 1$ and $1^1 = 1$ and $(2!)^2 = 4$ and $2^2 = 4$ so this inequality is only true for $n\ge 3$. ($(3!)^2 = 6^2 =36$ while $3^3 =27$)]

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The thing that strikes me is that $(n!)^2$ is a product of $n$ things: $k^2$ for $k=1....n$ and $n^n$ is a product of $n$ things: $n$.

If somehow the $i^2$ "average" to more than $n$ we'd have it.

But my second insight (although I don't see it immediately) is that we can pair the small small values of $1$ to $n$ with the large values of $n$ to $1$ to get $(n!)^2$ is the product of $n$ things: $k$th term from the bottom times $k$th term from the top.

That is:

$(n!)^2 = (1*2*.....*n)*(n*(n-1)*.... *1) = \prod_{k=1}^n k*(n+1 -k)$.

So if we can she $k*(n+1-k) > n$ we'd have it:

$(n!)^2 =\prod_{k=1}^n k*(n+1-k) > \prod_{k=1}^n n = n^n$

Does it?

$k*(n+1 - k) = kn + k - k^2 > n$ if

$(k-1)n + k > k^2$ if

$(k-1)n > k^2 -k$ if

$(k-1)n > k(k-1)$

If $k = 1$ then $k(n+1-k) = n$.

If $k > 1$ then this holds if

$n > k$. So if $n=k$ then $k(n+1-k) = n$.

but for all $1 < k < n$ we have $k(n+1-k) > n$.

So if $n=1$ we have $(n!)^2 =\prod_{k=1}^1 k*(2-k) = \prod_{k=1} 1=n^n$.

And if $n=2$ we have $(n!)^2 = \prod_{k=1}^2 k(3-k) =\prod_{k=1}^2 2 = n^2$.

But for $n\ge 3$ we have $(n!)^n = \prod_{k=1}^nk(n+1-k) = [1*n]*[\prod_{k=2}^{n-1} k(n+1-k)]*[n*1] > n*[\prod_{k=2}^{n-1} n]*n=\prod_{k=1}^nn = n^n$.

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tl;dr

If $n \ge 3$ and $1 < k < n$ then $k*([n+1]-k)=(k-1)k +kn> (k-1)n+kn = n$, while if $k=1$ or $k=n$ then $k*([n+1]-k)=n$.

So $\prod_{k=1}^n k*([n+1]-k) > \prod_{k=1}^n n$ or

$(\prod_{k=1}^n k)*(\prod_{j=n;-1}^1 j) > n^n$ or

$(n!)*(n!) = (n!)^2 > n^n$.