5
$\begingroup$

There are really lot of problems on AM-GM inequality because of its elementary nature and powerful applications.

What I want is a collection of questions/problems which look very complex but get solved swiftly and powerfully through use of AM-GM Inequality.

Edit:I see that 'interesting' is subjective hence to make this question more specific please provide examples which seem interesting or remarkable to you and don't think about what's interesting to me.

$\endgroup$
  • $\begingroup$ I think a good example are those optimisation exercises over the unit sphere which can be solved elegantly with mean inequalities. $\endgroup$ – Surb Dec 19 '14 at 16:39
  • $\begingroup$ Since remarkable and interesting are subjective criteria, it would probably be better for you to do your own search through Questions here that involve AM-GM inequality in their Answers. That way you take responsibility for collecting the ones that fit your ideas. $\endgroup$ – hardmath Dec 19 '14 at 16:39
  • $\begingroup$ An example: math.stackexchange.com/questions/839433/… $\endgroup$ – Surb Dec 19 '14 at 16:42
  • $\begingroup$ Here's a search of the kind I suggested. Variations may give you better results. $\endgroup$ – hardmath Dec 19 '14 at 16:45
  • 1
    $\begingroup$ No thanks are needed. I prefer to give you the tools to do your own work, believing that is how learning takes place. $\endgroup$ – hardmath Dec 19 '14 at 16:47
2
$\begingroup$

One cute application is to the well-known lemma:

Lemma: Let $\{x_n\}$ be a sequence of positive real numbers. Then $$\prod_{n=1}^{\infty} (1+x_n)$$ converges if and only if $$\sum_{n=1}^{\infty} x_n$$ converges.

Proof: One direction is clear, as $\prod_{n=1}^{N} (1+x_n) > \sum_{n=1}^{N} x_n$. For the other direction, use the AM-GM inequality. $$\sqrt[N]{\prod_{n=1}^{N} (1+x_n)}\le \frac{\sum_{n=1}^{N} (1+x_n)}{N} \implies \prod_{n=1}^{N} (1+x_n) \le \left(1 + \frac{x_1 + \cdots + x_N}{N} \right)^N$$ where the right hand side converges to $\exp{\sum_{n=1}^{\infty} x_n}$.

The usual way of proving this lemma is using $1+x\le e^x$ to get that upper bound. For some reason, I find it easy to forget this inequality, so AM-GM is a good fallback.

$\endgroup$
  • $\begingroup$ You can use that $t\leqslant \exp({t-1})$. $\endgroup$ – Pedro Tamaroff Dec 19 '14 at 17:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.