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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?

Connor Harris
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John Darson
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  • AM-GM gets you a lower bound of $(1^2 2^2 \cdots n^2)^{1/n}$, not $(1^2 2^2 \cdots n^2)^{n/2}$. – Connor Harris Jun 04 '18 at 21:51
  • The inequality is false for $n=1$. – uniquesolution Jun 04 '18 at 21:52
  • The AM-GM most certainly does not get you $(1^2 + 2^2 + ...n^2)/n > 1^22^2... n^2)^{n/2}$. Just try that nearly any valu $(1 + 4 + 9)/3 =14/3$ but $(1^22^23^3)^{3/2} = (123)^3 = 6^3$. Surely you don't believe $4\frac 13 > 216$!. – fleablood Jun 04 '18 at 22:01

1 Answers1

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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 $n=1$.

user
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