# Inequality over positive integers

Let $$x_1, x_2, \dots, x_n$$ and $$y_1, y_2,\dots, y_n$$ be two sets of positive integers such that $$x_i and $$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.

Thanks!

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 $$\sum_{i=1}^n x_i \left( y_i - \sqrt{x_n y_n} \right) \geq 0.$$
• 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. Feb 7 at 18:04