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

[johnqwertyful]

I have proven to n=3 an inequality that seems useful. x₁ⁿ+...+x_nⁿ≥nx₁...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?

 

[pwsnafu]

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]

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]

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]

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]

Take careful notice of where the constant n is used in the inequality:

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

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 x_1,x_2,x_3,x_4 since the inequality tells us to go up to x_n=x_4.

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

5^4+6^4+20^4+31^4 \geq 4\cdot 5\cdot 6\cdot 20\cdot 31

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

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

10^4+11^4+12^4+13^4 \geq 4\cdot 10\cdot 11\cdot 12\cdot 13

Don't sweat making the mistake though, all these x_n'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]

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,M_0(x_1,x_2,...,x_n) \le M_n(x_1,x_2,...,x_n) 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]

So in general, if w_1,...,w_n\in [0,1] with w_1+...+w_n=1, then we have

x_1^{n\cdot w_1}\cdot ...\cdot x_n^{n\cdot w_n} \leq w_1x_1^n + ... + w_nx_n^n

The original inequality has w_i=1/n.

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

w_1\varphi(y_1)+...+w_n\varphi(y_n)\leq \varphi\left(w_1y_1 + ... + w_ny_n\right)

The original inequlaity follows with \varphi = \log and y_i = x_i^n.

 

[dodo]

Ah, I see... if, with weights w_i = 1/n you were to choose an arbitrary power, say y_i = x_i^k, then in order to make \sqrt[n]{x_1^k x_2^k x_3^k ...} = x_1 x_2 x_3 ... at that point is where you would need k=n.

in other words, with a different power the inequality would look likex_1^k + x_2^k + ... + x_n^k \ge n \sqrt[n]{x_1^k x_2^k ... x_n^k}
Thanks for a very informative post, micromass... (but now I have a lot to read :)

 

[johnqwertyful]

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.