# Find constants $0<x_i\leq y_i<\infty$ such that conditions hold

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

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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

$$x_1>0,\quad x_2>0,\quad 0 $$\frac{\sqrt{x_1}}{\sqrt{x_2}\sqrt{x_3}}\leq y_1<\frac{1}{x_2 x_3},$$ $$x_2 $$x_3 < y_3\leq \frac{1}{x_2 y_1}.$$ I verified that this is indeed a solution.

Now, I am wondering if this solution can be generalized for $$n>3$$.

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

A special choice of constants would be to set $$x_i=x$$, $$y_i=y$$ for all $$i$$.
Then the desired conditions reduce to $$0 This can be reduced easily to $$y>0\\\begin{cases}01\end{cases}$$