Suppose that $x_{1},x_{2},\cdots,x_{n}$ are real numbers, such that $$x_{1}x_{2}\cdots x_{n}\neq 0$$ and $$\dfrac{x_{1}}{x_{2}}+\dfrac{x_{2}}{x_{3}}+\cdots+\dfrac{x_{n}}{x_{1}}=0$$

show that $$|x_{1}x_{2}+x_{2}x_{3}+\cdots+x_{n}x_{1}|\le \Bigg(\max_{1\le k\le n}|x_{k}|- \min_{1\le k\le n}|x_{k}|\Bigg)(|x_{1}|+|x_{2}|+\cdots+|x_{n}|)$$

This problem is from a math contest. I have already solved the following similar problem :(can see Mitrinovic D.S Analytic inequalitys Page 346 )

let $a_{i},x_{i}\in R$, such that $$\sum_{i=1}^{n}|x_{i}|=1,\sum_{i=1}^{n}x_{i}=0$$ show that :$$\Bigg|\sum_{i=1}^{n}a_{i}x_{i}\Bigg|\le\dfrac{1}{2}\Bigg(\max_{1\le i\le n}a_{i}-\min_{1\le i\le n}a_{i}\Bigg)$$ proof : Put $S_{k}=x_{1}+x_{2}+\cdots+x_{k}$. We assume wlog that $$a_{1}\ge a_{2}\ge\cdots\ge a_{n}$$ using Abel parts, we obtain $$I=\Bigg|\sum_{i=1}^{n}a_{i}x_{k}\Bigg|=\Bigg|S_{n}a_{n}+\sum_{k=1}^{n-1}S_{k}(a_{k}-a_{k+1})\Bigg|=\Bigg|\sum_{k=1}^{n-1}S_{k}(a_{k}-a_{k+1})\Bigg|$$ Now $|S_k|$ is smaller than $A=\sum_{i=1}^k |x_i|$, and smaller than $B=\sum_{i=k+1}^n |x_i|$. Since $A+B=1$, one of $A$ or $B$ is smaller than $\frac{1}{2}$. It is clear then that $$|S_{k}|\le \dfrac{1}{2}$$ so $$I\le \dfrac{1}{2}\sum_{k=1}^{n-1}(a_{k}-a_{k+1})=\dfrac{1}{2}(a_{1}-a_{n})$$

Perhaps a similar method applies here ?

My try: let $$\dfrac{x_{i}}{|x_{1}|+|x_{2}|+|x_{3}|+\cdots+|x_{n}|}=x'_{i}$$ so $$\sum_{i=1}^{n}|x'_{i}|=1$$ and $$\Longleftrightarrow \sum_{i=1}^{n}|x'_{i}x_{i}|\le (\max_{1\le k\le n}|x_{k}|- \min_{1\le k\le n}|x_{k}|)$$

But then I'm stuck. Thank you for any help

  • 1
    $\begingroup$ Interesting, some Chinese words up. $\endgroup$
    – Shine
    Aug 3, 2014 at 14:57
  • $\begingroup$ This simaler problem can see Mitrinovic D.S Analytic inequalitys Page 346 $\endgroup$
    – math110
    Aug 5, 2014 at 16:30

1 Answer 1


Let $Q$ be the product of $\max_{1\le k\le n}|x_{k}|$ and $\min_{1\le k\le n}|x_{k}|$, that is: $$Q := \max_{1\le k\le n}|x_{k}|\cdot\min_{1\le k\le n}|x_{k}|\,.$$

We can how represent each product of two neighboring (cyclically) numbers like $$ x_1 x_2 = Q\dfrac{x_1}{x_2} + x_1(x_2 - \dfrac{Q}{x_2})\,.$$

Then we have $$ x_1 x_2 + x_2 x_3 + \cdots + x_n x_1 = x_1(x_2 - \dfrac{Q}{x_2}) + x_2(x_3 - \dfrac{Q}{x_3}) + \cdots + x_n(x_1 - \dfrac{Q}{x_1}) $$ since $n$ terms of the type $Q\dfrac{x_{k-1}}{x_k}$ sum to zero due to the equality condition on our numbers. On the other hand, for each $l = 1\ldots n$ we have: $$ | x_l - \dfrac{Q}{x_l} | \le \max_{1\le k\le n}|x_{k}| - \min_{1\le k\le n}|x_{k}|\,,$$ that implies: $$ | x_{l-1}(x_l - \dfrac{Q}{x_l}) | \le |x_{l-1}|\Bigg(\max_{1\le k\le n}|x_{k}| - \min_{1\le k\le n}|x_{k}|\Bigg)\,.$$ When we sum all $n$ such inequalities, we’ll virtually obtain the inequality that has to be demonstrated (up to applying the triangle inequality for the absolute value function at the left-hand side).

  • $\begingroup$ Hello,can you explain why $$|x_{l}-\dfrac{Q}{x_{l}}|\le \max_{1\le k\le n}|x_{k}|-\min_{1\le k\le n}|x_{k}|?$$ $\endgroup$
    – math110
    Aug 11, 2014 at 14:42
  • $\begingroup$ For simplicity, let $\mathrm{Max} := \max_{1\le k\le n}|x_{k}|$ and $\mathrm{Min} := \min_{1\le k\le n}|x_{k}|$. So, $Q = \mathrm{Max}\cdot\mathrm{Min}$. Let’s prove that for any such $x$ that $\mathrm{Min} \le |x| \le \mathrm{Max}$ we have $| x - \dfrac{Q}{x} | \le \mathrm{Max} - \mathrm{Min}$. Okay, we have two cases (positive and negative $x$), but without loss of generalily may think that $x$ is positive, because $Q$ is positive and two terms under the difference ever have the same sign. We have: $$\mathrm{Min} \le \dfrac{Q}{x} \le \mathrm{Max}$$ , hence the inequality. $\endgroup$ Aug 11, 2014 at 15:27

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