Let $a_1,a_2,\dots,a_n>0$ be distinct positive real numbers and let $x_1,x_2,\dots,x_n \ge 0$ be non-negative real coefficients such that $\sum_{i=1}^n x_i = 1$.
Is it possible to find another set of non-negative real coefficients $y_1,y_2,\dots,y_n \ge 0, \sum_{i=1}^n y_i =1$ and $x_i \ne y_i$ for at least one $i$ such that $$ \sum_{i=1}^n x_i a_i = \sum_{i=1}^n y_i a_i? $$
For $n=2$ this is obviously impossible as $x_1 a_1 + x_2 a_2 = x_1 (a_1 - a_2) + a_2$ which is uniquely determined by $x_1$. However, I am not sure about higher $n$.
Any insights, references, or suggestions for further reading would be greatly appreciated.
Write $A = \sum_{i=1}^n a_i x_i$. This expression writes the number $A$ as a convex combination of the distinct positive reals $\{ a_i \}$. The question is equivalent to asking: given that there is at least one such expression, must there be another one?
Index the $a_i$ in increasing order WLOG, so that $a_1 < a_2 < \dots < a_n$. When $n = 2$ the set of convex combinations of $a_1$ and $a_2$ is just the line segment $[a_1, a_2]$, and every point in that line segment has a unique expression as a convex combination. So this is why the answer is no in this case.
When $n \ge 3$ the set of convex combinations is the line segment $[a_1, a_n]$ and the answer is still no if $A = a_1$ or $A = a_n$, since in both of these cases the unique convex combination is $x_1 = 1$ and all other $x_i$ zero or $x_n = 1$ and all other $x_i$ zero, respectively.
Otherwise, $A$ lies strictly between $a_1$ and $a_n$, and then the answer is yes. $A$ lies in the interval $[a_i, a_{i+1}]$ for some $1 \le i \le n-1$, so it can be expressed as a convex combination of $a_i$ and $a_{i+1}$. But since $A$ also lies in the larger interval $[a_1, a_n]$, it can also be expressed as a convex combination of $a_1$ and $a_n$, and because $n \ge 3$ this expression must be distinct from the previous one (since either $a_1 < a_i$ or $a_{i+1} < a_n$), unless $A = a_1$ or $a_n$ as previous.