# Does this inequality hold for two finite sequences?

Does this inequality hold?

Parameter conditions: $$a>0$$, $$x_i\in(-\frac{1}{a},\frac{1}{a})$$, $$y_i\geq 0$$, integer $$N\geq 2$$.

The inequality:

$$\displaystyle \frac{\frac{\sum_{i=1}^{N}y_i}{N^2}}{1-a^2(\frac{\sum_{i=1}^N x_i}{N})^2}\leq \frac{1}{N}\sum_{i=1}^N \frac{y_i}{1-a^2 x_i^2}$$

I have not found any simulation example to disprove this inequality. I have only found out under certain conditions, the inequality holds. I am wondering whether this inequality exists for general conditions as listed above.

First, the Cauchy-Schwarz Inequality asserts that: $$\left(\sum_{i=1}^N x_i\right)^2 \leq N\sum_{i=1}^N x_i^2$$ Therefore: \begin{align*} \frac{\frac{\sum_{i=1}^{N}y_i}{N^2}}{1-a^2\left(\frac{\sum_{i=1}^N x_i}{N}\right)^2} \leq \frac{\frac{\sum_{i=1}^{N}y_i}{N^2}}{1-a^2\frac{\sum_{i=1}^N x_i^2}{N}} = \frac{1}{N} \frac{\sum_{i=1}^N y_i}{N - a^2\sum_{i=1}^N x_i^2} = \frac{1}{N} \frac{\sum_{i=1}^N y_i}{\sum_{i=1}^N (1 - a^2x_i^2)} \end{align*} Since $$x_i \in \left(\frac{1}{a}, \frac{1}{a}\right) \implies 1 - a^2x_i^2 > 0$$, we have: $$\left(\sum_{i=1}^N \frac{y_i}{1 - a^2x_i^2}\right)\left(\sum_{i=1}^N (1 - a^2x_i^2)\right) \geq \sum_{i=1}^N y_i$$ The result follows.