Check the inequality of rational function with real numbers and positive weights

Question

I would like to know, if the following inequality can be true: $$\frac{\sum\limits_{i} p_i y_i -(\sum\limits_{i}p_i x_i)^2 }{(\sum\limits_{i}p_i z_i)^2} \leqslant \sum\limits_{i}p_i\left(\frac{y_i -x_i^2}{z_i^2}\right),$$ with $x_i, y_i, z_i \in R$ and positive weights $p_i>0$, such that $\sum_i p_i^2 = 1$. Also, $y_i - x_i^2 \geqslant 0$, for all $i$.The same holds for $\sum_ip_iy_i - (\sum_i p_ix_i)^2 \geqslant0$.

Answer 1

No it's not generally true.

By your extra conditions, both the LHS and the RHS are positive. Now, on the LHS, you can choose $z_i$ such that $(\sum_i p_i z_i)^2 \to 0_+$ which makes the LHS arbitrary large, hence the inequality won't hold.

Example: the sums run just over two values. $p_1 = p_2 = 1/\sqrt2$. $z_1=1$. $z_2 = -1 + \epsilon$. Then $(\sum_i p_i z_i)^2 = \epsilon^2/2$. As $\epsilon \to 0$, the denominator on the LHS gets arbitrarily small, whereas the denominators on the RHS vanish.