If, given a set of nonzero values, the ratio between the min and max is at least $\frac{1}{2}$, is this true for all other values inside the set? I haven't been able to prove this, or come up with a counterxample.
Given a set of nonzero values $A = \{a_1, a_2, ... a_n\}$, and given the hypothesis that ${\min(A)\over\max(A)}\geq\frac{1}{2}$, does the hypothesis imply that such condition is satisfied by every couple of elements $(a_i,a_j), i\neq j$?
I tried this by plugging in some numbers, and it appears to hold true, intuitively as well, but I haven't been able to generalize this yet.
Any help is highly appreciated.
 A: For all $k$ you have $\min(A)\leq a_k \leq \max(A)$. If $a_1,\ldots,a_n>0$, then for all $i,j$ it holds
$$\frac{a_i}{a_j} \geq \frac{\min(A)}{a_j}\geq \frac{\min(A)}{\max(A)}.$$
A: Take two distinct $a_i$ and $a_j$ from $A$. Then $a_i\ge\min(A)$ and $a_j\le\max(A)$. If $\min(A)>0$ then $a_j>0$ and $\max(A)>0$, so $\frac{1}{a_j}\ge\frac{1}{\max(A)}$. In that case $\frac{a_i}{a_j}\ge\frac{a_i}{\max(A)}\ge\frac{\min(A)}{\max(A)}$ (since we have $\frac{1}{\max(A)}>0$ and $a_i>0$). Then $\frac{a_i}{a_j}\ge\frac{1}{2}$.
If $\min(A)<0$ (we can't have $\min(A)=0$ by hypothesis) then we must have $\max(A)<0$, because in other case $\frac{\min(A)}{\max(A)}<0$, which can't happen since $\frac{\min(A)}{\max(A)}\ge\frac{1}{2}$. So in this case both $a_i$ and $a_j$ are negative numbers. In fact, once we have that $\max(A)<0$ it follows from $\max(A)\ge\min(A)$ that $\frac{\min(A)}{\max(A)}\ge1$. So your set has to be just of negative numbers, but taking that into account, every set you take will satisfy the inequality. However, it's easy to see that not every pair of negative numbers satisfy it, since we could take $\frac{-1}{-3}<\frac{1}{2}$ for example. So for $\min(A)<0$ the statement is false. It would be true if we require $a_i\le a_j$ in the fraction (or equivalentyly $|a_i|\ge|a_j|$, since they have to be negative numbers), by the same argument as before.
