Let $n$ be any integer greater than 1. Prove that
1^1\cdot 2^2\cdot 3^3\cdots n^n <\left(\frac{2n+1}{3} \right)^{\frac{n(n+1)}{2}}.
Solution. Let $N=1^1\cdot 2^2\cdot 3^3\cdots n^n$. Then there are
1+2+3+\cdots + n = \frac{n(n+1)}{2}
factors in $N$. Their geometric mean is $N^2/[n(n+1)]$ and their arithmetic mean is
$\begin{eqnarray}
&\frac{1+(2+2)+(3+3+3)+\cdots + (n+n+\cdots+n)}{\frac{n(n+1)}2}& \\
&= \frac{2(1+2^2+3^2+\cdots + n^2)}{n(n+1)}& \\
&= \frac{2n(n+1)(2n+1)}{6n(n+1)} = \frac{2n+1}{3}.&
\end{eqnarray}$
By the Arithmetic Mean - Geometric Mean Inequality, we have that $N^{2/[n(n+1)]}<(2n+1)/3$. That is,
N<\left(\frac{2n+1}3 \right)^{\frac{n(n+1)}2}.