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A conjecture

by johnqwertyful
Tags: conjecture
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johnqwertyful
#1
Feb16-13, 11:56 PM
P: 332
I have proven to n=3 an inequality that seems useful. x1n+...+xnn≥nx1...xn for all positive x.

I'm sure this has been proven before. I'm not quite sure how to extend it from n=3 to for all n. I'm thinking induction, but that has proven challenging. Any hints?
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pwsnafu
#2
Feb17-13, 01:43 AM
Sci Advisor
P: 820
This is a special case of the generalised mean inequality. You have coupled the number of variables with the index, they should be separate. Then you want to show that the mean is monotonically increasing in the index.
dodo
#3
Feb17-13, 04:15 AM
P: 688
Don't you require the x's to be >= 1 ? For example, 0.1^2 + 0.2^2 + 0.3^2 < 2*(0.1 + 0.2 + 0.3).

(I assume that, on the right-hand side, n multiplies the whole sum, which is not clear in the OP.)

P.S.: Actually, also 1.1^3 + 1.2^3 + 1.3^3 < 3*(1.1 + 1.2 + 1.3). So i'm at a bit of a loss.

Mentallic
#4
Feb17-13, 04:51 AM
HW Helper
P: 3,515
A conjecture

Quote Quote by Dodo View Post
Don't you require the x's to be >= 1 ? For example, 0.1^2 + 0.2^2 + 0.3^2 < 2*(0.1 + 0.2 + 0.3).

(I assume that, on the right-hand side, n multiplies the whole sum, which is not clear in the OP.)

P.S.: Actually, also 1.1^3 + 1.2^3 + 1.3^3 < 3*(1.1 + 1.2 + 1.3). So i'm at a bit of a loss.
The RHS isn't a sum, it's a product.
dodo
#5
Feb17-13, 05:05 AM
P: 688
Ah... thanks, Mentallic.

Still, 10^3 + 11^3 + 12^3 + 13^3 < 3*10*11*12*13 < 4*10*11*12*13.

(Not sure yet if the RHS multiplier is to be the power or the number of terms - though I can't find a counterexample if both are actually coupled.)
Mentallic
#6
Feb17-13, 05:21 AM
HW Helper
P: 3,515
Take careful notice of where the constant n is used in the inequality:

[tex]x_1^n+x_2^n+...+x_{n-1}^n+x_n^n \geq n\cdot x_1x_2...x_{n-1}x_n[/tex]

So what this means you should do is to take some value n, say, n=4, and now you need to choose 4 real positive numbers because we need to assign a value to [itex]x_1,x_2,x_3,x_4[/itex] since the inequality tells us to go up to [itex]x_n=x_4[/itex].

So if we used [itex]x_1 = 5, x_2 = 6, x_3 = 20, x_4 = 31[/itex] then we'd have

[tex]5^4+6^4+20^4+31^4 \geq 4\cdot 5\cdot 6\cdot 20\cdot 31[/tex]

If we used n=10 then we'd need to go up to [itex]x_{10}[/itex] and the LHS will use powers of 10.

Quote Quote by Dodo View Post
Ah... thanks, Mentallic.

Still, 10^3 + 11^3 + 12^3 + 13^3 < 3*10*11*12*13 < 4*10*11*12*13.


(Not sure yet if the RHS multiplier is to be the power or the number of terms - though I can't find a counterexample if both are actually coupled.)
This would work if you had

[tex]10^4+11^4+12^4+13^4 \geq 4\cdot 10\cdot 11\cdot 12\cdot 13[/tex]

Don't sweat making the mistake though, all these [itex]x_n[/itex]'s can be quite daunting at first sight, and unless they were clearly explained to you which it doesn't seem like they were, you're going to have a tough time guessing what they mean.
At least, I know I used to!
dodo
#7
Feb17-13, 05:39 AM
P: 688
Ah, thanks again. Then the power and the number of terms *need* to be coupled. Using the namings in the wiki page for the generalized mean inequality,[tex]M_0(x_1,x_2,...,x_n) \le M_n(x_1,x_2,...,x_n)[/tex] because 0 < n. Then raise each side to the n-th power to remove the n-th roots.

Though I can't say I understand the inequality, or know how to prove it, being the first time I see it. :)
micromass
#8
Feb17-13, 06:06 AM
Mentor
micromass's Avatar
P: 18,038
So in general, if [itex]w_1,...,w_n\in [0,1][/itex] with [itex]w_1+...+w_n=1[/itex], then we have

[tex]x_1^{n\cdot w_1}\cdot ...\cdot x_n^{n\cdot w_n} \leq w_1x_1^n + ... + w_nx_n^n[/tex]

The original inequality has [itex]w_i=1/n[/itex].

Even more general, we have Jensen's inequality that states that for any function [itex]\varphi:(0,+\infty)\rightarrow \mathbb{R}[/itex] such that [itex]\varphi^{\prime\prime}(t)\leq 0[/itex] for all t, then

[tex]w_1\varphi(y_1)+...+w_n\varphi(y_n)\leq \varphi\left(w_1y_1 + ... + w_ny_n\right)[/tex]

The original inequlaity follows with [itex]\varphi = \log[/itex] and [itex]y_i = x_i^n[/itex].
dodo
#9
Feb17-13, 07:12 AM
P: 688
Ah, I see... if, with weights [itex]w_i = 1/n[/itex] you were to choose an arbitrary power, say [itex]y_i = x_i^k[/itex], then in order to make [itex]\sqrt[n]{x_1^k x_2^k x_3^k ...} = x_1 x_2 x_3 ...[/itex] at that point is where you would need [itex]k=n[/itex].

in other words, with a different power the inequality would look like[tex]x_1^k + x_2^k + ... + x_n^k \ge n \sqrt[n]{x_1^k x_2^k ... x_n^k}[/tex]
Thanks for a very informative post, micromass... (but now I have a lot to read :)
johnqwertyful
#10
Feb17-13, 11:17 AM
P: 332
Wow, this has been an extremely helpful thread. Thanks everyone! I knew I wasn't the first to think of this. It's definitely a useful inequality.

I really appreciate the help, I'll look over the wiki page.


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