Proving an Inequality for All n: x1n+...+xnn≥nx1...xn

  • Thread starter johnqwertyful
  • Start date
  • Tags
    Inequality
In summary: Thanks again everyone!In summary, the conversation discusses a proven inequality that is a special case of the generalised mean inequality. The conversation also mentions the use of induction and the need for the x-values to be greater than or equal to 1. Further discussion delves into the use of weights and the relationship between the power and number of terms in the inequality.
  • #1
johnqwertyful
397
14
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?
 
Mathematics news on Phys.org
  • #2
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.
 
  • #3
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.
 
Last edited:
  • #4
Dodo said:
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.
 
  • #5
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.)
 
  • #6
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.

Dodo said:
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!
 
Last edited:
  • #7
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. :)
 
  • #8
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].
 
Last edited:
  • #9
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 :)
 
Last edited:
  • #10
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.
 

1. How do you prove an inequality for all n?

To prove an inequality for all n, you need to use mathematical induction. This involves showing that the statement is true for n = 1, and then assuming it is true for n = k and using that to prove it is true for n = k+1. This process is repeated until you have proven the statement for all n.

2. What is the significance of using x1n+...+xnn≥nx1...xn in inequalities?

This form of inequality is known as the Power Mean Inequality and is used to compare the arithmetic mean, geometric mean, and harmonic mean of a set of numbers. It is significant because it helps us understand the relationship between these different types of means and when they may be greater or less than each other.

3. Can you provide an example of using x1n+...+xnn≥nx1...xn to prove an inequality?

Sure, let's say we want to prove that for all positive integers n, the following inequality holds: x1n+x2n+...+xnn≥n * x1x2...xn. We can use mathematical induction to prove this. For n = 1, the statement becomes x1≥x1, which is true. Now, assuming the statement is true for some k, we have x1k+1+x2k+1+...+xk+1k+1≥(k+1) * x1x2...xk+1. Multiplying both sides by xk+1, we get x1k+1+x2k+1+...+xk+1k+1≥(k+1)x1x2...xk+1, which is true. Therefore, the statement is true for all positive integers n.

4. Are there any other important inequalities that can be proven using x1n+...+xnn≥nx1...xn?

Yes, there are many other important inequalities that can be proven using this method. Some examples include the Cauchy-Schwarz inequality, the AM-GM inequality, and the Rearrangement inequality. These are all useful tools in mathematics and can be proven using the Power Mean Inequality.

5. What are the applications of proving an inequality for all n using x1n+...+xnn≥nx1...xn?

The applications of proving an inequality for all n using this method are vast. Inequalities are used in various fields of mathematics, such as calculus, linear algebra, and number theory. They are also used in real-world applications, such as in economics, physics, and engineering. By understanding and proving these inequalities, we can better analyze and solve problems in these areas.

Similar threads

Replies
2
Views
1K
  • General Math
Replies
3
Views
2K
Replies
2
Views
2K
Replies
35
Views
3K
  • Calculus and Beyond Homework Help
Replies
6
Views
852
  • General Math
Replies
7
Views
3K
  • General Math
Replies
3
Views
1K
Replies
1
Views
1K
  • General Math
Replies
2
Views
2K
Back
Top