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I am getting crazy with this one.

Suppose $a_n=(1^2+2^2+3^2+\ldots+n^2)^n$ and $b_n=n^n(n!)^2$. Show that $a_n>b_n$ for all $n$.

They suggest to use the AM-GM inequality.

Mike Pierce
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Gian Luca
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3 Answers3

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From the desired inequality, $b_n$ should be the GM and $a_n$ the AM. To make an AM out of $a_n$, we should take the $n$th root and then divide by $n$. After this, the desired result becomes $$\frac{1^2+2^2+\ldots+n^2}{n} \stackrel?>\sqrt[n]{(n!)^2}$$ But by reordering the factors, indeed $(n!)^2=1^22^2\cdots n^2$ and we are done.

Hagen von Eitzen
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  • Thank you. I usually struggle so much in finding the hint for solving this kind of problems, and I believe you gave me a great hint. Thanks! – Gian Luca May 04 '15 at 19:08
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$$a_n> \left(n\cdot ((n!)^2)^{1/n}\right)^n=n^n(n!)^2=b_n$$ by AM-GM inequality. The first inequality follows from the AM-GM inequality since $$\frac{\displaystyle\sum_{i=1}^n i^2}{n}> \left(\prod_{i=1}^n i^2 \right)^{1/n}$$Note that the strict inequality follows since in AM-GM inequality involving sums and products of $a_i$, equality occurs iff $a_i$ are all same.

Samrat Mukhopadhyay
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    It would probably help the understanding a lot if you more explicit state how you are using the AM-GM inequality. – TravisJ May 04 '15 at 17:23
  • Also, note that the Question asks for strict inequality, though it should be straightforward to address that point. – hardmath May 04 '15 at 17:43
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AM-GM gives you $$ \frac{1}{n}\sum_{k=1}^{n} k^2 > \left( \prod_{k=1}^n k^2 \right)^{1/n} = (n!)^{2/n}, $$ because the $k^2$ are not all equal.

Chappers
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