Proof that $(1^2 + 2^2 + \cdots + n^2)^n > n^n (n!)^2$

Question

I need to prove that $a_{n} = (1^{2} +2^{2} + ...+n^{2})^{n} >n^{n}(n!)^{2}=b_{n}$ for any natural number n.

Proof:

Note that by the arithmetic-geometric mean inequality,$\frac{1^{2} +2^{2} + ...+n^{2}}{n} > (1^{2}\cdot 2^{2}\cdot...\cdot n^{2})^{n/2}$. Then $1^{2} +2^{2} + ...+n^{2} > n(1^{2}\cdot 2^{2}\cdot...\cdot n^{2})^{n/2} = n(n!)^{n}$. This implies that $a_{n} =(1^{2} +2^{2} + ...+n^{2})^{n} >n^{n}(n!)^{2n} \geq n^{n}(n!)^{n}= b_{n}$. Then we conclude that $a_{n} > b_{n}$ for all natural numbers $n$.

Is there anything wrong with my proof?

Answer 1

It is correct but by AM-GM we have

$$\frac{1^{2} +2^{2} + ...+n^{2}}{n} \ge (1^{2}\cdot 2^{2}\cdot...\cdot n^{2})^{1/n} $$

from which we conclude

$$\implies (1^{2} +2^{2} + ...+n^{2})^n\ge n^n(n!)^2$$

and equality holds only for the case $n=1$.