Prove by induction that $1^1\cdot 2^2\cdot \dots\cdot n^n\leq \left(\frac{2n+1}{3}\right)^{\frac{n(n+1)}{2}}.$ How can I prove by induction that for every natural number $n$, 
$$1^1\cdot 2^2\cdot \dots\cdot n^n\leq \left(\frac{2n+1}{3}\right)^{\frac{n(n+1)}{2}}.$$
 A: Hint. Instead of induction, I warmly recommend the use of the AGM inequality. Note that
$$\left(1^1\cdot 2^2\cdot \dots\cdot n^n\right)^{\frac{2}{n(n+1)}}$$
is the geometric mean of the numbers
$$1,2,2,3,3,3,\dots,\underbrace{n,\dots,n}_{\text{$n$ times}}.$$
What is their arithmetic mean?
$$\frac{1+2+2+3+3+3+\dots+\overbrace{n+\dots+n}^{\text{$n$ times}}}{n(n+1)/2}=\frac{1+2^2+3^2+\dots+n^2}{n(n+1)/2}=?$$
A: The inductive proof requires the Bernoulli inequality (using $3$ terms).
\begin{eqnarray*}
(2n+3)^{n+2}=(2n+1+2)^{n+2} \geq  \\(2n+1)^{n+2} +2(n+2) (2n+1)^{n+1} + \frac{(n+2)(n+1)}{2} 2^2 (2n+1)^{n}=(2n+1)^n (10n^2+20n+9) > (3(n+1))^2 (2n+1)^n.
\end{eqnarray*}
So we have $(2n+3)^{n+2}  \geq  (3(n+1))^2 (2n+1)^n$ & the result will follow with a bit of algebra.
The inductive step requires
\begin{eqnarray*}
1 \times 2^2 \cdots n^n  \times(n+1)^{n+1} \leq ( \frac{2n+1}{3})^{\frac{n(n+1)}{2}}  (n+1)^{n+1} \leq ( \frac{2n+3}{3})^{\frac{(n+1)(n+2)}{2}}
\end{eqnarray*}
So we need to show 
\begin{eqnarray*}
 ( \frac{2n+1}{2n+3})^{\frac{n(n+1)}{2}} \leq ( \frac{2n+3}{3(n+1)})^{(n+1)}
\end{eqnarray*}
Take the $n+1$ root and use the previously derived result.
