For what positive integers $n \ge 3$ inequality holds $$x_1^2+x_2^2+...+x_n^2 \ge p(x_1x_2+x_2x_3+...+x_{n-1}x_n)$$ for all real numbers, if $p=1; p=4/3; p=6/5$

My work so far:

a) $p=1 \Rightarrow $for all $n$: $$x_1^2+(x_1-x_2)^2+(x_2-x_3)^2+...+(x_{n-1}-x_n)^2+x_n^2 \ge 0$$

b) $p=4/3 \Rightarrow n \le 3$

$n=3:$ $$(3x_1-2x_2)^2+x_2^2+(2x_2-3x_3)^2\ge0$$ $n=4$. If $x_1=x_4=2; x_2=x_3=3 -$ inequality is wrong.

c) $p=6/5 \Rightarrow n \le 4$

$n=4:$ $$(3x_1-2x_2)^2+x_2^2+15(x_2-x_3)^2+x_3^2+(3x_3-5x_4)^2\ge0$$ $n=4$. If $x_1=x_5=9; x_2=x_4=15, x_3=16 -$ inequality is wrong.

How to investigate this problem for $p>0$?


1 Answer 1


The question is equivalent to asking whether the quadratic form given below is positive semidefinite. The form is $x^TAx$ for $$A=\left[\begin{matrix}1 & -p/2 & 0 & 0&\cdots & 0&0\\ -p/2 &1&-p/2 & 0& \cdots & 0&0\\ \vdots &&&&\ddots&&\vdots\\ 0 & 0 & 0& 0&\cdots& -p/2 &1\end{matrix}\right]$$ where $A$ has $1$ on the main diagonal, and $-p/2$ on the two diagonals immediately above and below.

This form is positive semidefinite if all of the eigenvalues of $A$ are nonnegative. Fortunately, the matrix $A$ is tridiagonal and Toeplitz, and therefore has particularly nice eigenvalues: $$1+2\sqrt{\frac{p^2}{4}}\cos(k\pi/(n+1)),\text{ for }k=1,2,\ldots, n$$ These terms are decreasing as $k$ increases, so we can say that they are all nonnegative if the last one is, that is if $$0\le 1+2\sqrt{\frac{p^2}{4}}\cos(n\pi/(n+1))$$ which we can rearrange as $$0\le 1+|p|\cos\left(\pi \frac{n}{n+1}\right)$$

In particular, if $|p|<1$, then the statement holds for all $n$, as cosine is at most one in absolute value. If $|p|>1$, then for $n$ sufficiently large the statement will fail to hold, as $\lim_{n\to\infty}\cos\left(\pi \frac{n}{n+1}\right)=-1$, and hence $$\lim_{n\to\infty}1+|p|\cos\left(\pi \frac{n}{n+1}\right)=1-|p|<0$$


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