Let $a_i\geq0$ for all $i\in\{1,2,...,n\}$ with $\sum_{i=1}^na_i=1$. Suppose $x_1,x_2,...,x_n$ are non-negative reals. Show that
$(\sum_{i=1}^na_i^2x_i^2)(\sum_{i=1}^n\dfrac{1}{x_i^2})\leq\dfrac{(z_{\min}+z_{\max})^2}{4z_{\min}z_{\max}}$ where $z_\min=\min_i(x_i^2a_i)$ and $z_\max=\max_i(x_i^2a_i)$
Since, $$\left(z_{max} - \frac{a_ix_i}{1/x_i}\right)\left(z_{min} - \frac{a_ix_i}{1/x_i}\right) \le 0 \implies z_{max}z_{min}\frac{1}{x_i^2} + a_i^2x_i^2\le a_i(z_{max}+z_{min})$$
Thus adding the inequalities together:
$$\displaystyle z_{max}z_{min}\sum\limits_{i=1}^{n}\frac{1}{x_i^2} + \sum\limits_{i=1}^{n}a_i^2x_i^2 \le (z_{max}+z_{min})\sum\limits_{i=1}^{n}a_i$$
Applying Am-Gm to the RHS,
$\displaystyle 2\sqrt{z_{max}z_{min}\sum\limits_{i=1}^{n}\frac{1}{x_i^2}. \sum\limits_{i=1}^{n}a_i^2x_i^2} \le z_{max}z_{min}\sum\limits_{i=1}^{n}\frac{1}{x_i^2} + \sum\limits_{i=1}^{n}a_i^2x_i^2 \le (z_{max}+z_{min})$
establishes the required inequality.
