# An inequality to prove

1. Nov 26, 2013

### DorelXD

1. The problem statement, all variables and given/known data

Let $0 ≤ x_1 ≤ x_2 ≤ ... ≤ x_n$ and $x_1 + x_2 + ... + x_n = 1$ , All the 'x' are real and n is a natural number. Prove the following:

$$(1+x_1^21^2)(1+x_2^22^2)...(1 + x_n^2n^2) ≥ \frac{2n^2+9n+1}{6n}$$

2. Relevant equations

3. The attempt at a solution

Well....all I managed to do was to "check the field a little". First of all, the right member of the inequality somehow reminds me of the sum:

$$1^2 + 2^2 + ... + n^2 = \frac{n(n+1)(2n+1)}{6} = \frac{(n+1)(2n^2 + 3n + 1)}{6}$$

I know from experience that we can solve this type of exercises by using some well-know inequalities such as the one that relates the average mean with the geometric one, or the Cauchy-Buniakowski-Schwarz ( C-B-S ). So:

$$1 ≥\frac{1}{n} ≥ \sqrt[n]{x_1x_2...x_n}$$ , so it follows immediately that : $$x_1x_2...x_n ≤ 1$$

That's all I could do. Could you help me continue, please? This exercise has been stuck in my head for two weeks already. I can't solve it alone, and I would like that somebody could guide me to the solution. Thank you!

Last edited: Nov 26, 2013
2. Nov 26, 2013

### Ray Vickson

Are all the $x_i \geq 0?$

3. Nov 26, 2013

### DorelXD

Yes, they are. I also modified the first post, by adding this fact.

4. Nov 27, 2013

### DorelXD

Can you help me ? Do you have any idea :D ?

5. Nov 27, 2013

### haruspex

This is under precalculus. If calculus methods are permitted I would use Lagrange multipliers.

6. Nov 27, 2013

### PAllen

You can make elementary (non calculus) arguments that the left side of the inequality is ≥ (1+n). Then, you can get the rest by algebra.

7. Nov 27, 2013

### Ray Vickson

One can solve the optimization problem
$$\min \sum_{i-1}^n \log(1 + i^2 x_i^2) \\ \text{subject to}\\ x_1 + x_2 + \cdots + x_n = 1 \\ 0 \leq x_1 \leq x_2 \leq \cdots \leq x_n$$
using a numerical optimization package for various small to medium values of n (3 ≤ n ≤ 10). In all cases I tried the optimal solution is $x_1 = x_2 = \cdots = x_n = 1/n$. Furthermore, this can be shown to be a so-called Kuhn-Tucker point for the problem with general n. The minimization objective is pseudo-convex and the constraint set is convex, so satisfying the Karush-Kuhn-Tucker conditions is, I believe, sufficient for a (global) minimum.

In other words, we have
$$(1+1^2 x_1^2)(1 + 2^2 x_2^2) \cdots (1+ n^2 x_n^2) \geq (1 + (1/n)^2)(1+(2/n)^2) \cdots (1 + (n/n)^2)$$
for all feasible x. Maybe that can be manipulated further to give the desired result.

8. Nov 27, 2013

### haruspex

This is confusing. 'n' is the number of terms, so the nth term means just the last term. Certainly xn >= 1/n, so the last term is at least 2. I don't see how you get it's at least 1+n.
If you meant a more generic n, I still don't see it. xj need not be ≥ 1/j, nor 1/n. Every xj except xn could be zero.

9. Nov 27, 2013

### PAllen

Sorry, I realized my mistake as soon as finished, and was working on deleting my post. I missed that the x's were squared as well as the i's. Oops.

10. Nov 27, 2013

### haruspex

Suppose xj < xj+1 for some j < n. It might be straightforward to show that substituting xj' = xj+1' = (xj + xj+1)/2 produces a lower product. (Calculus by the back door.)

11. Nov 27, 2013

### haruspex

Yeah, I've been there - a frantic dash to delete a regretted post, only to find it's already been read.

12. Nov 27, 2013

### PAllen

Unfortunately, this is not true. The substitution produces a larger product. Larger by the difference between xj and its successor, over 2, squared.

13. Nov 27, 2013

### haruspex

Hmm.. I just tried it and it worked for me. I ended up needing that 1 > j(j+1)xjxj+1, which is clearly true.

14. Nov 27, 2013

### Ray Vickson

A much easier way to get a lower bound is to note that for positive $y_i = i^2 x_i^2$ we have
$$(1+y_1)(1+y_2) \cdots (1+y_n) > 1 + \sum_{i=1}^n y_i,$$
because the RHS is just the first two terms in the expansion of the product, and all the omitted terms are > 0. Therefore, we can get a lower bound by solving the relatively simple problem
$$\min \sum_{i=1}^n i^2 x_i \:\:\:\leftarrow \sum_{i=1}^n y_i\\ \text{subject to}\\ \sum_{i=1}^n x_i = 1\\ 0 \leq x_1 \leq x_2 \leq \cdots \leq x_n$$
This problem has the optimal solution $x_1 = x_2 = \cdots = x_n = 1/n$, so we have
$$(1+1^2 x_1^2)(1+2^2 x_2^2) \cdots (1 + n^2 x_n^2) > 1 + \sum_{i=1}^n i^2/n^2$$
for all feasible $x_1, x_2, \ldots x_n$.

Last edited: Nov 27, 2013
15. Nov 27, 2013

### PAllen

I keep forgetting the squaring.

Last edited: Nov 27, 2013
16. Nov 27, 2013

### PAllen

This seems promising. This lower bound appears to equal the RHS. If this could be proven (along with the rest), you'd be done.

17. Nov 28, 2013

### Millennial

We need the proof for this in elementary terms.
If this is proven, the problem is basically solved.

Edit: I just saw that the $x_i$ had to be positive. Then, the proof is obviously very easy.
The sum $\displaystyle \sum_{i=1}^n i^2/n^2$ has closed form $\displaystyle \frac{(n+1)(2n+1)}{6n}$. The inequality then follows from the optimal solution.

Last edited: Nov 28, 2013