Let $x_1, x_2, \dots, x_n$ and $y_1, y_2,\dots, y_n$ be two sets of positive integers such that $$x_i<y_i \quad \text{for all } i$$ and $$x_iy_i \leq x_{i+1}y_{i+1} \quad \text{for all } i$$ I am studying when the inequality $$\sum_{i=1}^n x_i\left(y_i - \sqrt{x_ny_n}\right)\geq 0$$ holds.

It is clear that if $y_i\geq \sqrt{x_ny_n} \quad \text{for all } i$ then the inequality holds. Can we prove something stronger, like "if $\frac{\sum_{i=1}^n y_i}{n} \geq \sqrt{x_ny_n}$ then the inequality holds"? Or some other condition of this type.



1 Answer 1


Since $x_i < y_i$ and $x_i y_i < x_{i+1} y_{i+1}$, we can write $y_i > x_i$ and $x_{i+1}y_{i+1} > x_i y_i$, which implies that \begin{align} y_{i+1}^2 &= y_{i+1}y_{i+1} \\ &> x_{i+1}y_{i+1} \\ &> x_i y_i \\ &> x_ix_i \\ &= x_i^2 \end{align} for all $i= 1, \ldots, n$. So we have \begin{align} & \ \ \ \sum_{i=1}^n x_i \left( y_i - \sqrt{x_n y_n} \right) \\ &= \sum_{i=1}^n \left( x_i y_i - x_i \sqrt{x_n y_n } \right) \\ &= \sum_{i=1}^n x_i y_i - \sqrt{x_n y_n } \sum_{i=1}^n x_i \\ &\geq n x_1 y_1 - \sqrt{x_n y_n} \sum_{i=1}^n x_i \\ &> n x_1^2 - \sqrt{y_n^2} \sum_{i=1}^n x_i \\ &= n x_1^2 - y_n \sum_{i=1}^n x_i. \end{align} Thus if $$ \sum_{i=1}^n x_i \leq \frac{n x_1^2}{y_n}, $$ then we have $$ \sum_{i=1}^n x_i \left( y_i - \sqrt{x_n y_n} \right) \geq 0. $$

  • $\begingroup$ thanks for your answer! I like the sufficient condition for the inequality to hold that you have proved. $\endgroup$ Feb 7 at 17:51
  • $\begingroup$ Noting that $y_1 \geq x_1 + 1$ strengthens your condition a bit. If we have $\sum_{i=1}^n x_i \leq \frac{n x_1^2 +nx_1}{y_n}$ then the inequality holds. $\endgroup$ Feb 7 at 18:04

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