Induction Using AM-GM Inequality

Question

I have the following problem that I am stuck on.

Recall the AM-GM inequality: for set of distinct positive real numbers $x_{1},x_{2},\ldots,x_{n}$, we have $\frac{x_{1}+x_{2}+\cdots+x_{n}}{n}\geq\sqrt[n]{x_{1}x_{2}\cdots x_{n}}$. Suppose $a_{n}=(1^{2}+2^{2}+\cdots+n^{2})^{n}$ and $b_{n}=n^{n}(n!)^{2}$. Show that $a_{n}\geq b_{n}$ for all $n$.

I know that I clearly need to use induction. For the base case, we have $a_{1}=(1^{2})^{1}=1$ and $b_{1}=1^{1}(1!)^{2}=1$ so that $a_{1}\geq b_{1}$. Now I assume that the statement holds for some $k\in\mathbb{N}$ such that $(1^{2}+2^{2}+\cdots+k^{2})^{k}\geq k^{k}(k!)^{2}$. We must show that the statement holds for $k+1$. I keep getting stuck on the inductive step (perhaps I am missing something obvious?). Here's what I have so far:

$\begin{split} (1^{2}+2^{2}+\cdots+(k+1)^{2})^{k+1}&=(1^{2}+2^{2}+\cdots+(k+1)^{2})^{k}(1^{2}+2^{2}+\cdots+(k+1)^{2})\\ \\ &\geq (1^{2}+2^{2}+\cdots+k^{2})^{k}(1^{2}+2^{2}+\cdots(k+1)^{2})\\ \\ &\geq k^{k}(k!)^{2}(1^{2}+2^{2}+\cdots+(k+1)^{2}) \hspace{0.25cm} (\mathrm{by\:inductive\:hypothesis})\\ \\ &=k^{k}(k+1)(k!)^{2}\frac{1^{2}+2^{2}+\cdots+(k+1)^{2}}{k+1}\\ \\ &\geq k^{k}(k+1)(k!)^{2}\frac{1+2+\cdots+(k+1)}{k+1}\\ \\ &\geq k^{k}(k+1)(k!)^{2}\sqrt[k+1]{1\cdot 2\cdots(k+1)} \hspace{0.25cm} (\mathrm{by\:AM-GM\:inequality})\\ \\ &=k^{k}(k+1)(k!)^{2}[(k+1)!]^{1/(k+1)} \end{split}$

But from here, I have no idea how to show that this quantity is larger than $(k+1)^{k+1}[(k+1)!]^{2}$. Am I on the right track? Is there something simple that I'm missing? Thanks in advance for any help or suggestions!

Answer 1

Set $x_i=i^2$ in the AM-GM inequality.

Answer 2

No induction required here: observe the A.M.-G.M. inequality can be written as $$\Bigl(\frac{x_1+x_2+\dots+x_n}{n}\Bigr)^{\!n}\ge x_1x_2\dotsm x_n,$$ and the given inequality as $$\Bigl(\frac{1^2+2^2+\dots+n^2}{n}\Bigr)^{\!n}\ge (1\cdot 2\dotsm n)^2=1^2\cdot 2^2\dotsm n^2.$$