The assertion is false. A counterexample with $n=2$ is
$$
a_1=a_2=x_1=x_2=1,\quad b_1=\tfrac12,\, y_1=4,\, b_2=2,\, y_2=\tfrac14.
$$
Suppose we make the change of variables $w_i = a_i^2x_i$ and $z_i = b_i^2y_i$; then the assertion becomes
$$
\frac{\sum a_i^{-1}w_i}{\sum b_i^{-1}z_i} \le \frac{\sum w_i}{\sum z_i} = 1 \implies \frac{\sum a_iw_i}{\sum b_iz_i} \ge 1.
$$
In this formulation, it's natural to consider the case $w_i=z_i=1$; it's easy now to see that the assertion is likely false if we take the set of $a_i$ to be closed under taking reciprocals and the set of $b_i$ to be closed under taking reciprocals.
One can proceed similarly as in the proof of the “regular” rearrangement inequality: If $\sigma$ is a permutation of $\{1, \ldots ,n\}$ and not the identity then there are indices $j < k$ such that exchanging $\sigma(j)$ and $\sigma(k)$ gives a new permutation $\tau$ with more fixed points than $\sigma$ and
$$ \tag{*}
\sum_{i=1}^n f(x_i + y_{\sigma(i)}) \le \sum_{i=1}^n f(x_i + y_{\tau(i)}) \, .
$$
If $\tau$ is not the identity then this step can be repeated, and after finitely many steps one obtains
$$
\sum_{i=1}^n f(x_i + y_{\sigma(i)}) \le \sum_{i=1}^n f(x_i + y_i) \, .
$$
In the case of the “regular” rearrangement inequality one uses that for $a_1 \le a_2$ and $b_1 \le b_2$
$$
(a_2-a_1)(b_2-b_1) \ge 0 \implies a_1 b_2 + a_2 b_1 \le a_1 b_1 + a_2 b_2 \, .
$$
In our case one can use the following to prove $(*)$:
If $f$ is a convex function and $a_1 \le a_2$ and $b_1 \le b_2$ then
$$
f(a_1 + b_2) + f(a_2 + b_1) \le f(a_1 + b_1) + f(a_2 + b_2) \, .
$$
This holds trivially if $a_1 =a_2$ or $b_1 = b_2$. In the case $a_1 < a_2$ and $b_1 < b_2$ it follows from adding the convexity conditions:
$$
f(a_1 + b_2) \le \frac{a_2-a_1}{a_2+b_2-a_1-b_1} f(a_1 + b_1) + \frac{b_2 - b_1}{a_2+b_2-a_1-b_1} f(a_2 + b_2) \\
f(a_2 + b_1) \le \frac{b_2-b_1}{a_2+b_2-a_1-b_1} f(a_1 + b_1) + \frac{a_2 - a_1}{a_2+b_2-a_1-b_1} f(a_2 + b_2)
$$
For positive sequences $u_1, \ldots, u_n$ and $v_1, \ldots, v_n$ the normal rearrangement inequality follows from the generalized one with $f(t)=e^t$ applied to $x_i = \log u_i$ and $y_i = \log v_i$, since then
$$
f(x_i + y_{\sigma(i)}) = u_i \cdot v_{\sigma(i)} \ .
$$
It is also a consequence of Karamata's inequality: Set
$$
(a_1, a_2, \ldots , a_n) = (x_n + y_n, x_{n-1}+y_{n-1}, \ldots, x_1 + y_1)
$$
and let $(b_1, b_2, \ldots , b_n)$ be a decreasing rearrangement of
$$
(x_n + u_n, x_{n-1}+u_{n-1}, \ldots, x_1 + u_1) \, .
$$
Then
$$
(a_1,a_2,\ldots,a_n)\succ(b_1,b_2,\ldots,b_n)
$$
so that
$$
f(a_1)+f(a_2)+ \ldots +f(a_n) \ge f(b_1)+f(b_1)+ \ldots +f(b_n)
$$
which is the desired conclusion.
Best Answer
let $y_i = r_i x_i$ and let $r_{min}$ and $r_{max}$ be the smallest and largest values of $r$, respectively.
Then for all $y_i$, $r_{min}x_i \le y_i \le r_{max}x_i$ as $x_i > 0$
This gives $r_{min}\sum_{i=1}^n x_i \le \sum_{i=1}^n y_i \le r_{max}\sum_{i=1}^n x_i$ so $$r_{min} \le \frac{\sum_{i=1}^n y_i}{\sum_{i=1}^n x_i} \le r_{max}$$
Which means either ${(\frac{\sum_{i=1}^n y_i}{\sum_{i=1}^n x_i})}^2 \le {r_{min}}^{2}$ or ${(\frac{\sum_{i=1}^n y_i}{\sum_{i=1}^n x_i})}^2 \le {r_{max}}^{2}$
But ${r_{min}}^{2}$ and ${r_{max}}^{2}$ are both terms in the RHS sum. This means there is always a single term greater than the LHS, so the whole sum on the RHS must be greater than the LHS. In fact, it means the inequality is strict for $n>1$.
Edit: It is strict if there is more than one $y_i$ not equal to $0$. As Martin R correctly said.