Hidden Cauchy-Schwarz inequality

Question

If $x_{i}\gt 0$ and $x_{i}y_{i}-{z_{i}}^2\gt 0$ for $i\le n$, then prove that $$\frac{n^3}{(\sum_{i=1}^nx_{i})(\sum_{i=1}^ny_{i})-(\sum_{i=1}^n{z_{i}}^2)} \le \sum_{i=1}^n\frac 1{x_{i}y_{i}-z_{i}^2}$$ This is a question from the IMO 1969. At first it seems that we can assume that $a_{i}=\sqrt{x_{i}y_{i}}-z_{i}$ and $b_{i}=\sqrt{x_{i}y_{i}}+z_{i}$. Hence RHS becomes $\sum_{i=1}^n\frac 1{a_{i}b_{i}}$. Now how do we simplify the LHS so that we can use the Cauchy-Schwarz inequality?

Answer 1

let, $f_{i}(x)=x_{i}x^2-2z_{i}x+y_{i}$.

Hence, $\min\limits_{x\in\mathbb R}f_{i}(x)=\frac{x_{i}y_{i}-z_{i}^2}{x_{i}}$.

$\min\limits_{x\in\mathbb R}\sum\limits_{i=1}^{n}f_{i}(x)\geq\sum\limits_{i=1}^{n}\min\limits_{x\in\mathbb R}f_{i}(x)$.

Hence, $\frac{\sum\limits_{i=1}^nx_{i}\cdot\sum\limits_{i=1}^{n}y_{i}-(\sum\limits_{i=1}^{n}z_{i})^2}{\sum\limits_{i=1}^{n}x_{i}}\geq\sum\limits_{i=1}^{n}\frac{x_{i}y_{i}-z_{i}^2}{x_{i}}$ and

$\left( \sum\limits_{i=1}^{n}x_{i}\cdot \sum_{i=1}^{n}y_{i}-\left( \sum\limits_{i=1}^{n}z_{i}\right)^{2}\right)\cdot\sum\limits_{i=1}^{n}\frac{1}{x_{i}y_{i}-z_{i}^{2}}\geq$

$\geq\sum\limits_{i=1}^{n}x_{i}\cdot\sum\limits_{i=1}^{n}\frac{x_{i}y_{i}-z_{i}^{2}}{x_{i}}\cdot\sum\limits_{i=1}^{n}\frac{1}{x_{i}y_{i}-z_{i}^{2}}\geq n^{3}.$