A problem on inequality and number theory

Question

Suppose $n \in \mathbb{Z}^+$;

$$\sum_{i=1}^n\sqrt{1-x_ix_{i+1}}\ge\sqrt{n(n-1)}\tag{1}$$

where $, x_1= x_{n+1}, \displaystyle \quad\forall x_i\in\mathbb{R} \text{ and }\sum_{i=1}^nx_i^2=1 $

Is it possible to find all the positive integer numbers $n$ which satisfy $(1)$ ?

I cannot make head or tails of this. Obviously neither (1) nor (2) holds for all $x_i\in\Bbb{R}$, which is what you are saying. Are you asking for a list of integers $n$ such that both (1) and (2) hold for some choice of $x_i$s?? Anyway, this ain't number theory. Were you just throwing tags at random without understanding what they mean? — Jyrki Lahtonen, Oct 08 '16 at 08:10
I think, but the asker will have to clarify, that we want to find all natural numbers $n$ such that (2) holds whenever (1) does. — Dylan, Oct 08 '16 at 08:14
That's another possibility, @Dylan. Possibly a more interesting one:-) We need to wait for the asker to edit the question to make it clear what they mean. — Jyrki Lahtonen, Oct 08 '16 at 08:16
Yes, (1) looks more like a nonlinear constraint to the nonlinear optimization like problem (2). — user6043040, Oct 08 '16 at 08:16
I would vote to close as unclear what you are asking, but because my vote is instantly binding that is not kosher. — Jyrki Lahtonen, Oct 08 '16 at 08:17
I hope it is now clear after editing. thanks for comments and suggestions! — user6043040, Oct 08 '16 at 08:25
It's still not entirely clear if you want to find all $n$ such that we can choose the $x_i$'s so that the inequality holds, or if you want to find all $n$ such that no matter how we choose the $x_i$'s, the inequality always holds as long as $\sum x_i^2 = 1$. (I think that the second version is the intended one, because the answer for the first version is "all $n$".) — Dylan, Oct 08 '16 at 08:35
It was said that, for $n>20$ that is no such solution, and for $n\le 5$ the proof would be not so difficult. Post the answers here, I can give you at least my upvote :) — user6043040, Oct 08 '16 at 09:58

score 2 · Accepted Answer · answered Oct 08 '16 at 10:43

A proof for $n=4$.

Id est, for $a^2+b^2+c^2+d^2=1$ we need to prove that: $$\sqrt{1-ab}+\sqrt{1-bc}+\sqrt{1-cd}+\sqrt{1-da}\geq2\sqrt3$$ or $$\sum\limits_{cyc}(1-ab)+2\sum\limits_{cyc}\sqrt{(1-ab)(1-bc)}+2\sqrt{(1-ab)(1-cd)}+2\sqrt{(1-bc)(1-da)}\geq12$$ or $$\sum\limits_{cyc}\sqrt{(1-ab)(1-bc)}+\sqrt{(1-ab)(1-cd)}+\sqrt{(1-bc)(1-da)}\geq4+\frac{1}{2}\sum\limits_{cyc}ab$$ Now by C-S we obtain $$\sum\limits_{cyc}\sqrt{(1-ab)(1-bc)}=\sum\limits_{cyc}\sqrt{(a^2-ab+b^2+c^2+d^2)(b^2-bc+c^2+a^2+d^2)}=$$ $$=\sum\limits_{cyc}\sqrt{\left(\left(b-\frac{a}{2}\right)^2+\frac{3}{4}a^2+\frac{3}{4}c^2+\frac{1}{4}c^2+d^2\right)\left(\left(b-\frac{c}{2}\right)^2+\frac{3}{4}a^2+\frac{3}{4}c^2+\frac{1}{4}a^2+d^2\right)}\geq$$ $$\sum\limits_{cyc}\left(\left(b-\frac{a}{2}\right)\left(b-\frac{c}{2}\right)+\frac{3}{4}a^2+\frac{3}{4}c^2+\frac{1}{4}ac+d^2\right)=$$ $$=\frac{7}{2}-ab-bc-cd-da+ac+bd$$ In another hand, by C-S again we obtain $$\sqrt{(1-ab)(1-cd)}=\sqrt{(a^2-ab+b^2+c^2+d^2)(c^2-cd+d^2+a^2+b^2)}=$$ $$=\sqrt{\left(\frac{(a-b)^2}{2}+\frac{1}{2}+\frac{1}{2}c^2+\frac{1}{2}d^2\right)\left(\frac{(c-d)^2}{2}+\frac{1}{2}+\frac{1}{2}a^2+\frac{1}{2}b^2\right)}\geq$$ $$\geq\frac{(a-b)(d-c)}{2}+\frac{1}{2}+\frac{1}{2}(ad+bc)=\frac{1}{2}+ad+bc-\frac{1}{2}(ac+bd)$$ similarly we'll obtain $$\sqrt{(1-ad)(1-bc)}\geq\frac{1}{2}+ab+cd-\frac{1}{2}(ac+bd)$$ Thus, $$\sum\limits_{cyc}\sqrt{(1-ab)(1-bc)}+\sqrt{(1-ab)(1-cd)}+\sqrt{(1-bc)(1-da)}\geq\frac{9}{2}$$ and it remains to prove that $$\frac{9}{2}\geq4+\frac{1}{2}\sum\limits_{cyc}ab$$ or $$a^2+b^2+c^2+d^2\geq ab+bc+cd+da$$ or $$(a-b)^2+(b-c)^2+(c-d)^2+(d-a)^2\geq0$$ Done!

This is already the best results so far so I accept this answer though there is only result for $n=4$. Since this post seems unclear, I created another here: http://math.stackexchange.com/questions/1985923/inequality-problems-in-different-cases-of-n-difficult — user6043040, Oct 27 '16 at 01:33

A problem on inequality and number theory

1 Answers1