Let $n\geq 2$ and suppose $x_1,x_2,\ldots,x_n$ and $y_1,y_2,\ldots,y_n$ are constants such that, for each $i=1,2,\ldots,n$, one has
$$
0<x_i < y_i<\infty.
$$
Edit
I would like to know if (and if yes: how) it is possible to choose the constants such that
$$
\prod_{j=1}^n y_j\leq\frac{y_i}{x_i}\leq\left(\prod_{j=1}^nx_j\right)^{-1}\qquad\textrm{for all }i=1,2,\ldots,n.\tag{1}
$$
I think it makes sense to start with $n=2$ and $n=3$.
For $n=2$, (1) holds if we choose
$$
x_2:=y_1^{-1},\quad y_2:=x_1^{-1}.
$$
However, already the case $n=3$ seems to be much more difficult.
For $n=3$, Wolfram Alpha says that one possible solution is
$$
x_1>0,\quad x_2>0,\quad 0<x_3<\frac{1}{x_1 x_2},
$$
$$
\frac{\sqrt{x_1}}{\sqrt{x_2}\sqrt{x_3}}\leq y_1<\frac{1}{x_2 x_3},
$$
$$
x_2<y_2\leq\frac{1}{x_3 y_1},
$$
$$
x_3 < y_3\leq \frac{1}{x_2 y_1}.
$$
I verified that this is indeed a solution.
Now, I am wondering if this solution can be generalized for $n>3$.
Best Answer
A special choice of constants would be to set $x_i=x$, $y_i=y$ for all $i$.
Then the desired conditions reduce to $$ 0<x<y\\ y^n\leq y/x\leq x^{-n} $$ This can be reduced easily to $$ y>0\\\begin{cases}0<x<y&y\leq 1\\0<x\leq y^{1-n}&y>1\end{cases} $$