Let $x_1\leq x_2\leq ... \leq x_n$ and $y_1<y_2< ... <y_n$ be positive integers such that $x_1\geq 2$, $x_i < y_i$ and $y_i + 1<y_{i+1}$. Do we have that $$x_ny_n (\sum_{i=1}^n x_i)^2 \leq (\sum_{i=1}^n x_iy_i)^2$$

The proposed inequality has a certain "Cauchy-Schwarz" flavour, but from the lower bound side of the sum of products of two variables. Given the conditions, the right side has the potential to grow more rapidly due to both the squaring of the sum and the strict increase of $y_i$​ elements. However, I feel that proving this inequality requires a sophisticated mathematical argument or method that I am not able to develop.

The conditions have been set after finding counterexamples if they are softened. However, if you think that they can be softened in any way, it would be great. Nevertheless, here my


(I) I have not been able to find counterexamples to the inequality with the conditions set. Do you find any? If there is, how would you strengthen the conditions for the inequality to hold?

(II) Do you see feasible obtaining a proof of the inequality? Any hint would be welcomed.

(III) Could you share some reference of similar "lower bounds with Cauchy-Schwarz" flavour? It would be great to know about inequalities of this type.



After @NeckverseHerdman counterexample, I have refined my searching and found that the following condition seems to be sufficient: $$x_ny_n\leq \frac{\left(\sum_{i=1}^n y_i\right)^2}{n^2}$$ But then, the initial inequality becomes $$\left(\sum_{i=1}^n x_i\right)^2 \cdot \left(\sum_{i=1}^n y_i\right)^2 \leq n^2 \cdot \left(\sum_{i=1}^n x_iy_i\right)^2$$ $$\left(\sum_{i=1}^n x_i\right) \cdot \left(\sum_{i=1}^n y_i\right) \leq n \cdot \left(\sum_{i=1}^n x_iy_i\right)$$ This seems more feasible to prove using some kind of rearrangement theorem, any hint would be welcomed. THANKS!

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    $\begingroup$ I made a little program in Python to loop over some integer sets to give me counterexamples and I found a lot of them! Perhaps one of the easiest is the following: Take $n = 2$ and for $x_1,x_2$ you can take $2$ and $4$ and for $y_1, y_2$ you can take $3$ and $5$. This satisfies the conditions but gives us a counterexample to the inequality: $$x_2 y_2 \left(x_1 + x_2 \right)^2 = 4 \times 5 \times (6)^2 = 20 \times 36 = 720$$ while $$(x_1y_1 + x_2 y_2)^2 = (6 + 20)^2 = 26^2 = 676.$$ $\endgroup$ Commented Mar 2 at 22:17
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    $\begingroup$ @NeckverseHerdman thanks for the counterexample. It seems that I need to add another condition for the inequality to hold. $\endgroup$ Commented Mar 3 at 12:04

1 Answer 1


I think I can prove that for every choice of real numbers $x_1\leq x_2\leq \dots \leq x_n$ and $y_1\leq y_2\leq \dots \leq y_n$ we have that $$\left(\sum_{i=1}^n x_i\right) \cdot \left(\sum_{i=1}^n y_i\right) \leq n \cdot \left(\sum_{i=1}^n x_iy_i\right)$$ It can be done with a direct application of the rearrangement inequality, which states that $x_1y_{\sigma(1)}+ \dots +x_ny_{\sigma(n)}\leq \sum_{i=1}^n x_iy_i$ for every permutation $\sigma$ of the numbers $1,2,...,n$. If we sum the $n$ sums of permutations where each $x_i$ is impaired with every $y_i$ is precisely $\left(\sum_{i=1}^n x_i\right) \cdot \left(\sum_{i=1}^n y_i\right)$, and by the rearrangement inequality, as each of those sums of permutations is equal or less than $\sum_{i=1}^n x_iy_i$, we get the desired inequality. An inequality no doubt more beautiful and general than the one stated originally.

Regarding the original inequality, we have that, for every choice of real numbers $x_1\leq x_2\leq \dots \leq x_n$ and $y_1\leq y_2\leq \dots \leq y_n$ such that $x_ny_n\leq \frac{\left(\sum_{i=1}^n y_i\right)^2}{n^2}$ we have that $$x_ny_n (\sum_{i=1}^n x_i)^2 \leq (\sum_{i=1}^n x_iy_i)^2$$

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    $\begingroup$ That is Chebyshev's sum inequality $\endgroup$
    – Martin R
    Commented Mar 4 at 8:53
  • $\begingroup$ @MartinR thanks for the reference! I did not know about it, and yes, it is the same inequality, so great to know it is correct $\endgroup$ Commented Mar 4 at 10:09

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