Show using inequality of means that $a\cdot n \cdot \frac{1}{n} \le a^2n^2+\frac{1}{n^2}$

Question

Show using inequality of means that for $a>0$ and $n\in\mathbb{N}$: $$a\cdot n \cdot \frac{1}{n} \le a^2n^2+\frac{1}{n^2}$$

I'm sure it's not that complicated, but I'm probably missing something..

This is actually a small part of a proof which I want to understand.

user2345215 · Accepted Answer · 2015-01-25 16:30:07Z

2

This is a simple case of AM-GM and we can do even better: $$a^2n^2+\frac1{n^2}\ge2\sqrt{a^2n^2/n^2}=2a$$

Alternatively, $$0\le\left(an-\frac1n\right)^2=a^2n^2-2a+\frac1{n^2}\implies a^2+\frac1{n^2}\ge 2a$$

answered Jan 25, 2015 at 16:30

user2345215

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$\begingroup$ Thank you. btw, I missed an $1/2$, but it doesn't really matter I guess $\endgroup$
– AlonAlon
Jan 25, 2015 at 16:31
$\begingroup$ (+1) Even if $a \le 0$, $a^2n^2+\frac{1}{n^2} - a = \frac{3}{4}(an-\frac{1}{n})^2+\frac{1}{4}(an+\frac{1}{n})^2$ :) $\endgroup$
– sciona
Jan 25, 2015 at 22:11
1

$\begingroup$ @sciona Thanks. In that case $((an-\frac1n)^2+a^2n^2+1/n^2)/2$ seems more natural to me, but I guess that's just a personal opinion :) $\endgroup$
– user2345215
Jan 25, 2015 at 22:27

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sciona · Accepted Answer · 2015-01-25 22:11:58Z

2

In fact the given inequality is sharp. It follows from AM-GM inequality:

$\dfrac{1}{2}\left(a^2n^2+\dfrac{1}{n^2}\right) \geq \sqrt{a^2n^2.\dfrac{1}{n^2}}=a.n.\dfrac{1}{n}$

sciona

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answered Jan 25, 2015 at 16:30

voldemort

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