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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 $n>3$.

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  • $\begingroup$ All constants equal to one is always a solution, isn't it? $\endgroup$
    – maxmilgram
    Nov 16, 2023 at 7:23
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    $\begingroup$ Please edit the question accordingly. $\endgroup$
    – maxmilgram
    Nov 16, 2023 at 9:23

1 Answer 1

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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 $$ y>0\\\begin{cases}0<x<y&y\leq 1\\0<x\leq y^{1-n}&y>1\end{cases} $$

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