Given positive integers $k<n$ and positive real numbers $x_1$, $x_2, \dots, x_n$. Denote $$ A={x_1\over x_2+x_3+\dots+x_{k+1}}+{x_2\over x_3+x_4+\dots+x_{k+2}}+\ldots+{x_n\over x_1+x_2+\dots+x_k}$$

Prove that a) $A\geq \dfrac{n}{k^2}$; b) $A\geq \log 2\cdot \dfrac{n}{k}$.

This is a problem from a Russian contest. For the second part I have no idea, and for the part, I was trying to apply Cauchy-Schwarz but it didn't work out. I think I made some mistake, because the first part seems workable with CS. This was my attempt :

$$A=\sum_{\text{cyc}}\dfrac{x_1^2}{x_1x_2+x_1x_3+\cdots+x_1x_{k+1}}\ge \dfrac{(x_1+x_2+\dots+x_n)^2}{P}$$ Now this $P$ is confusing me. What is it? And even if I get it, does it imply this is greater than $\dfrac{n}{k^2}?$

Hope someone can help me. Thanks a lot.

Also something which made me think, is there a better constant, than $\log 2$ in the inequality?

  • $\begingroup$ use CS for the denominator first. $\endgroup$
    – DeepSea
    Jul 19, 2014 at 16:09
  • $\begingroup$ @8pir Could you clarify your comment? I don't see how it works, thanks a lot. $\endgroup$
    – shadow10
    Jul 19, 2014 at 16:11
  • $\begingroup$ What about setting $x_1 = \dots = x_n = 1$ then $A = \frac{n}{k}$. $\endgroup$
    – cactus314
    Jul 22, 2014 at 2:06
  • $\begingroup$ Yes, @johnmangual that gives the best constant is $\le 1$. But, does it show anything about the inequality I gave? I mainly need to look for the expression $P$ in my attempt. Thanks. $\endgroup$
    – shadow10
    Jul 22, 2014 at 14:32
  • $\begingroup$ Are you sure there is no more conditions on $ x_i $'s? $\endgroup$ Jul 23, 2014 at 3:50

1 Answer 1


As you have used Cauchy-Schwarz Inequality on, $$A=\dfrac{x_1}{x_2+x_3+\dots+x_{k+1}}+\dfrac{x_2}{x_3+x_4+\dots+x_{k+2}}+\ldots+\dfrac{x_n}{x_1+x_2+\dots+x_k}$$

leads to: $\displaystyle A \ge \dfrac{(x_1+\cdots+x_n)^2}{\sum\limits_{i=1}^{k}x_i(x_{i+1}+x_{i+2}+\cdots+x_{i+k})}$ where the summation is taken on cyclic index ($\displaystyle x_i = x_{i \pmod n}$).

So, $\displaystyle P = \sum\limits_{i=1}^{k}x_i(x_{i+1}+x_{i+2}+\cdots+x_{i+k})$ and $S = x_1+\cdots+x_n$ (say).Then, $\displaystyle A \ge \frac{n}{k^2}$ but even so, $\displaystyle \dfrac{S^2}{P}$ need not be greater than $\displaystyle \frac{n}{k^2}$.

As for the interpretation of $P$, it is a Quadratic Form with matrix of the form being a Symmetric circulant matrix the first row is given by writing a row of $0$'s and adding in succession $\displaystyle \frac{1}{2}$ in positions $2,3,\cdots, k+1$ and in the positions $n-k+1,\cdots,n$ of the row (with possible modification due to a previous addition if $2m \ge n$). The rest of the rows are cyclic permutations of row 1.

Now, $\displaystyle S^2 - \frac{n\lambda}{k}P$ (for some $\lambda > 0$) is a Quadratic Form and it is $\ge 0$ iff the matrix of the form is positive-semi-definite, i.e., iff its eigenvalues are non-negative.

The first row of the matrix of the form $\displaystyle S^2 - \frac{n\lambda}{k}P$ is given by is given by writing a row of $1$'s and subtracting in succession $\displaystyle \frac{n\lambda}{2k}$ in positions $2,3,\cdots, k+1$ and in the positions $n-k+1,\cdots,n$ of the row (with possible modification due to a previous subtraction if $2m \ge n$). The eigenvalues (of circulant matrix) being $$\lambda_j = \sum\limits_{r=1}^{n} \omega_j^r - \frac{n\lambda}{2k}\sum\limits_{r=1}^{k} (\omega_j^r + \omega_j^{-r})$$ for $j=0,1,2,\cdots,n-1$, and $\omega_j = e^{2\pi ij/n}$.

Clearly, $\displaystyle \lambda_0 = n - \dfrac{n\lambda}{2k}.2k = n(1-\lambda) \ge 0$ iff $\lambda \le 1$.

and, $\displaystyle \lambda_j = \dfrac{n\lambda}{2k}\dfrac{\sin \frac{j\pi}{n} - \sin (2k+1)\frac{j\pi}{n}}{\sin \frac{j\pi}{n}}$ for $j \ge 1$. So, the form is positive(semi-definite) depending on whether: $$\displaystyle \sin \frac{j\pi}{n} \ge \sin (2k+1)\frac{j\pi}{n}$$ for all $j = 1,2,\cdots, n-1$ (independent of $\lambda$) which one can certainly see is not true for all $k,n$.

As for $\displaystyle A \ge \dfrac{n}{k^2}$, take $x_{i_1} = \max\{x_1,x_2,\cdots,x_n\}$

and define $x_{i_2} = \max\{x_{i_1+1},x_{i_1+2},\cdots,x_{i_1+k}\}$,

till $x_{i_m} = \max\{x_{i_{m-1}+1},x_{i_{m-1}+2},\cdots,x_{i_{m-1}+k}\}$

and $x_{i_{m+1}} = x_{i_1}$.

Then clearly, $m > \dfrac{n}{k}$.

Also, $\displaystyle A \ge \dfrac{x_{i_1}}{x_{i_1+1}+x_{i_1+2}+\cdots+x_{i_1+k}} + \dfrac{x_{i_2}}{x_{i_2+1}+x_{i_2+2}+\cdots+x_{i_2+k}} + \cdots + \dfrac{x_{i_m}}{x_{i_m+1}+x_{i_m+2}+\cdots+x_{i_m+k}} \ge \dfrac{x_{i_1}}{kx_{i_2}}+\dfrac{x_{i_2}}{kx_{i_3}}+\cdots + \dfrac{x_{i_m}}{kx_{i_1}} \ge \dfrac{m}{k} \ge \dfrac{n}{k^2}$

by Am-Gm Inequality.

The second inequality looks non trivial. The case $k=2$: $\displaystyle \sum\limits_{i=1}^{n} \dfrac{x_i}{x_{i+1}+x_{i+2}} \ge \frac{n\lambda}{2}$.

The case $\lambda = 1$ is known as the Shapiro-Inequality which does not hold for all $n$.

The best constant $\lambda_{max} = 0.9891..$ as mentioned in the Wiki-link.


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