Finding the maximum possible number of elements of a set of real numbers Let $0<b < a$ be two given real numbers. Consider a set of real numbers which satisfies the following conditions:


*

*Each number in the set is strictly greater than $0$ but less than or equal to $b$.

*The summation of all numbers in the set is equal to $a$.

*The summation of each pair of numbers in the set is strictly greater than $b$.
What can we tell about the maximum number of elements of such a set? For example, for $a=2$ and $b=1$ at most $3$ elements are possible to assign to such a set (e.g., $\{0.5,0.7,0.8\}$ or $\{0.2,0.9,0.9\}$ etc.).

We can order the numbers $x_i$ in this set as
$$0<x_1\leq x_2 \leq\ldots\leq x_n\leq b\ .$$
Since the $x_i$ sum to $a$ we have
$$(n-1)x_2\leq x_2+x_3+\ldots+x_n=a-x_1\ ,$$
hence 
$$n\leq{a-x_1\over x_2}+1\ .$$
The condition $x_1+x_2>b$ implies $-x_1<x_2-b$ and $x_2>{b\over2}$. We therefore obtain
$$n<{a-b+x_2\over x_2}+1={a-b\over x_2}+2<{a-b\over b/2}+2={2a\over b}\ .$$
On the other hand the resulting bound $n_*=\lceil 2a/b\rceil-1$ is optimal and can be attained with equal pieces of size slightly larger than ${b\over2}$.
