- #1

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_{1}

^{n}+...+x

_{n}

^{n}≥nx

_{1}...x

_{n}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?

- Thread starter johnqwertyful
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- #1

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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?

- #2

pwsnafu

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- #3

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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.

(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.

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- #4

Mentallic

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The RHS isn't a sum, it's a product.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.

- #5

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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.)

- #6

Mentallic

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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.

[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!

[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.

This would work if you had

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.)

[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!

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- #7

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Though I can't say I understand the inequality, or know how to prove it, being the first time I see it. :)

- #8

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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].

[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].

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- #9

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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 :)

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 :)

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- #10

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

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