Suppose I have a set of $N$ real numbers $\{x_i\}$ and I have in general, for each $i$, that $x_i^2\leq x_i$. Suppose also that I have that $\sum_i x_i^2 = \sum_i x_i$. I want to show that $x_i$ = $x_i^2$ for each $i$. How would I go about doing that rigorously (it's intuitively clear me)? I know it's going to be something so simple (do I maybe have to use a multinomial expansion?.
This is not a homework question. This came up as part of a proof in Ballentine's Quantum Mechanics: A Modern Development.
Suppose there were an index $j$ with $x_j^2 < x_j$. Then we would have $$x_j-x_j^2 >0$$ and $x_i-x_i^2 \ge 0$ for all other indices. Hence $$0 = \sum_i \left ( x_i - x_i^2 \right ) \ge (x_j-x_j^2) >0,$$ which is a contradiction. Hence $x_i = x_i^2$ for all indices $i$.