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We have to prove that $$\frac{(x_1+x_2+x_3+...+x_n)}{n} \geq (x_1\cdot x_2\cdot x_3\cdots x_n)^{1/n}$$

Attempt: Raising both sides to the nth power gives $\left(\frac{x_1+x_2+x_3+...x_n}{n}\right)^{n} \geq x_1x_2x_3...x_n$ This is equivalent to proving that if a sum of random numbers is a fixed number, then their maximum product occurs when all the number are equal. So we have to find which combination of numbers gives the maximum product if their sum is fixed.

Obviously $$\left({x_1+x_2+x_3+...x_n}\right)^{n} > x_1x_2x_3...x_n$$

then, a maximizer must exist.

Now suppose that we have this combination of products: $x_1x_2x_3...x_n$ If i take the arithmetic mean of any 2 numbers and replace them by their arithmetic mean, I will get a bigger product (By the simple 2 case AM-GM inequality which is easy to proof), obviously the arthimetic mean of 2 numbers will not change the sum of the numbers, we can write that:

$$\frac{x_1+x_2}{2}\cdot\frac{x_1+x_2}{2}\cdot x_3 \cdots x_n\ \geq x_1x_2x_3\cdots x_n$$

Which means if we have any combination of product of numbers, I could increase that product by making any 2 numbers equal (arthimetic mean) without changing the sum of the numbers.

Now supposing that the maximum product does not have all of the numbers equal, then I can increase its product by having 2 numbers equal, so this mean that it's not the maximum product, then, by contradiction, the maximum product must occur when all the numbers are equal (Sum divided by the number of numbers)

Which means that $$\left(\frac{x_1+x_2+x_3+...x_n}{n}\right)^{n} \geq x_1x_2x_3...x_n$$

Is this a valid proof? It's my first time here and I don't know how to do the math notations thingy on this website so forgive me, thanks.

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    $\begingroup$ meta.math.stackexchange.com/questions/5020/… $\endgroup$ – Timbuc Jun 21 '15 at 9:59
  • $\begingroup$ I fixed the LaTeX (math notation). Please look to my edit and to the tutorial in the link of Timbuc's comment, to learn it. $\endgroup$ – wythagoras Jun 21 '15 at 10:01
  • $\begingroup$ It seems mathematically correct to me, though I would expand on why (x1+x2...xn)^n is greater than x1x2...xn. I assume you use the binomial formula to show this? $\endgroup$ – Faraz Masroor Jun 21 '15 at 14:14
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    $\begingroup$ Thanks wythagoras, @FarazMasroor We have let $x_{1}+x_{2}+...x_n$ be a fixed number, so let's name it $M$. Since $M$ is bigger than any $x_n$, $M\cdot M\cdot M\cdot M...$ (n times) must be bigger than $x_1\cdot x_2\cdot x_3...x_n$ (n times) $\endgroup$ – Wejd Jun 22 '15 at 16:53
  • $\begingroup$ That is another way to do it! $\endgroup$ – Faraz Masroor Jun 22 '15 at 21:55
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That's pretty good.

One aspect I'd want to fiddle with is the existence of the maximizer. You argued that it's bounded (which as a commenter noted could be made more explicit) and so a maximizer exists; this requires some kind of compactness argument, and the continuity of $\mathbb R^n\to\mathbb R$, $(x_1,\dotsc,x_n)\to\prod_{k=1}^n x_k$. That's all fine, but it would be better to make it explicit.

If you don't want to use topological concepts like that, then the argument can be adjusted not to assume the existence of the maximizer, but to prove it. The idea is to repeatedly replace pairs of $x_i$ until they're all the same, and argue (as you did) that we can do this keeping the AM constant but increasing the GM. The only tricky bit is that, if you replace pairs of numbers with their average, this process might not ever stop. (Example: $(1,2,2)\to (\frac32,\frac32,2)\to (\frac32,\frac74,\frac74) \to\dotsm$.) So the replacement step has to be tweaked. The resulting proof can be found in a note by Dijkstra, EWD1140.

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  • $\begingroup$ That's really interesting, I actually thought about the continuity of the process,(Replacing 2 numbers in the set with their arithmetic mean), each time you do this process you increase the product,and since the product is bounded, then, this process must "stop" somewhere which means that it must converge, and if it converges,then the product of the previous process compared to the next process must be equally the same somewhere in infinity, thus if i take the mean of any 2 numbers, the product must not change, then those 2 numbers must be equal, and this is satisfied if all of them are equal $\endgroup$ – Wejd Jun 22 '15 at 17:09
  • $\begingroup$ I did not really understand Dijkstra's proof but it's good to know the name of the problem $\endgroup$ – Wejd Jun 22 '15 at 17:11
  • $\begingroup$ Perhaps instead of replacing the numbers with the average you replace one with the average of all the numbers, and the other accordingly to keep the sum constant. $\endgroup$ – Faraz Masroor Jun 22 '15 at 21:57
  • $\begingroup$ @FarazMasroor Indeed, that is what Dijkstra does. $\endgroup$ – user21467 Jun 22 '15 at 23:50
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Hint: Your idea is correct. Here is a north to let things rigorous.

Consider $f, \psi : U \to \mathbb R$ defined as $$ f(x) = x_1 \cdot x_2 \cdots x_n \,\,\, \text{and} \,\,\, \psi (x) = x_1 + x_2 + \ldots + x_n$$

for all $x = (x_1, x_2, \ldots , x_n) \in U$. Fix $s > 0$ and search for the critical points of $f|_M$ where $M = \psi^{-1}(s)$.

Observe that $\mathrm{grad} \, \psi (x) = (1,1,\ldots, 1)$ for any $x \in U$ and $\mathrm {grad}\, f (x) = (\alpha_1 , \ldots , \alpha_n)$, with $a_i = \prod_{j\neq i} x_j$. Show that the maximum is necessarily in $M$ and is the only critical point of $f|_M$.

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  • $\begingroup$ And the downvote is because...? $\endgroup$ – Aaron Maroja Jun 21 '15 at 14:24
  • $\begingroup$ Other users than the downvoter themself can only speculate what was the reason for the downvote. But if you wish to discuss the reason for the downvote, there is a chatroom explicitly for this purpose. $\endgroup$ – Martin Sleziak Jun 21 '15 at 16:44
  • $\begingroup$ The problem is, the approach is correct, though people will look at this and think otherwise. $\endgroup$ – Aaron Maroja Jun 21 '15 at 16:59
  • $\begingroup$ No idea who downvoted, I know one can prove it through gradients with your approach :) but I tried to make the proof having mathematics as simple as possible $\endgroup$ – Wejd Jun 22 '15 at 16:33
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    $\begingroup$ Well, not in English ;) $\endgroup$ – user940 Jul 1 '15 at 22:53

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