Combinatorics: Prove that $x, y, z$ exist such that $\frac {1}{2} \leq \frac {x^2}{yz} \leq 2$

Question

Suppose we have a subset of $2n-1$ numbers from the set ${1, 2, 3, ..., 2^n-2}$. Prove that there exits $x, y, z$ such that $\frac {1}{2} \leq \frac {x^2}{yz} \leq 2$

I'm fairly new to this topic, and i honestly have no idea how to prove this. Any hint on how can i start the proof?

Thanks in advance.

Answer 1

HINT.- Consider the rational function $f(X)=\dfrac{(X-2)^2}{(X-3)(X-4)}$.

It is easy to prove that $$\begin{cases}f(X)\le2\text{ when } X\gt7\\f(X)\ge\dfrac12\text{ when } X\gt5\end{cases}$$ Making now $X=2^x$ the inequalities above prove that for $n\ge3$ (in which case $X\ge8$) a solution is given by $$x=2^n-2\\y=2^n-3\\z=2^n-4$$ It remains to solve the values $n=1$ and $n=2$ in which cases $(x,y,z)=(1,1,1),(1,1,2)$ are trivial solutions.