Proof by induction, factorials and exponents I've spent many hours trying to find proof by induction to this problem:
$$(n!)^2<\left(\frac{(n+1)(2n+1)}{6}\right)^n$$
for n > 1.
Does anyone have any idea how to solve it? Any help would be appreciated. 
 A: Using AM-GM inequality (which can be proved by induction on the number of terms),
$$\sqrt[n]{1^2\cdot 2^2\cdot3^2 \cdots n^2} \le \frac{1^2 + 2^2+3^2+\cdots + n^2}{n}$$
Equality holds iff $1^2 =  2^2 =  \ldots = n^2$, which means equality does not hold for $n>1$.
On the left hand side,
$$(n!)^2 = 1^2 \cdot2^2 \cdot 3^2 \cdots n^2$$
which can be proved by induction on $n$.
On the right hand side,
$$1^2 + 2^2+3^2+\cdots + n^2 = \frac{n(n+1)(2n+1)}{6}$$
which can also be proved by induction on $n$.
Joining the three links together,
$$\sqrt[n]{(n!)^2} < \frac{(n+1)(2n+1)}{6}$$
Taking the $n$th power on both sides (which preserves order as both sides are positive) gives the required inequality.
A: I will prove the given inequality with a different approach from @peterwhy just to shopw another way I like to prove this.  It is a little bit more euler- handwavy than @peterwhy s proof but it hink it is quite nice.
to start we note two well known results (which I can prove if you want)
$an^n>bn!>c d^n$ for all n suitably large in relation to a, b, c and d and only for n suitably large. This can be summarized as $n^n$ grows faster than $n!$ which grows faster than $d^n$. 
proving this is not hard.
for this problem we are interested if $an^n> c d^n bn!$ for all n suitably large in relation to a, b, c and d and only for n suitably large. say we have $e=\frac{cb}{a}$ we take n so large that lets say $q=(n!)^{1/n}$ then $q^n=n!$ notice that $q<n$ and that $n-q$ can be made arbitrarly large. then we can say that $d^n n!=(dq)^n$ if we then pick $n=(e^{1/n}dq)$ , $an^n= c d^n bn$ while for every $n$ smaller than that $an^n< c d^n bn!$ and for every $n$ larger $an^n> c d^n bn$ because $\frac{ c d^n bn!}{a}=\frac{ cb}{a}(dq)^n=e(dq)^n=(e^{1/n}dq)^n$.
After proving this we notice that $(n!)^2<\left(\frac{(n+1)(2n+1)}{6}\right)^n$ implies $6^n(n!)^2<\left((n+1)(2n+1)\right)^n>(n^2)^n=(n^n)^2$ since we know that $(n^n)^2>6^n(n!)^2$ because $(n^n)>6^n(n!)$ for all n suitably large and only for n suitably large. we only have to show that the original inequality holds for $n=2$ which it does.
