Romania — NMO Selection Tests for the Balkan and International Mathematical Olympiads

Let
$$f(x_1, x_2, \dots, x_n) = \max \left\{ \frac{x_k}{1 + x_1 + x_2 + \dots + x_k} : k = 1, 2, \dots, n \right\},$$
where $x_1 \ge 0, x_2 \ge 0, \dots, x_n \ge 0$ and $x_1 + x_2 + \dots + x_n = 1$, and notice that
$$\frac{a_1}{1+a_1} = \frac{a_2}{1+a_1+a_2} = \dots = \frac{a_n}{1+a_1+a_2+\dots+a_n}$$
for a unique $n$-tuple $(a_1, a_2, \ldots, a_n)$ of non-negative real numbers which add up to 1,
namely, $a_k = 2^{k/n} - 2^{(k-1)/n}$, $k = 1, 2, \ldots, n$, in which case $f(a_1, a_2, \ldots, a_n) = 1 - 2^{-1/n}$. Now let $(x_1, x_2, \ldots, x_n) \neq (a_1, a_2, \ldots, a_n)$, where the $x_i$ are non-
negative real numbers which add up to 1. Since $x_1+x_2+\cdots+x_n = a_1+a_2+\cdots+a_n$,
it follows that $x_k > a_k$ for some index $k$. Let $m = \min\{k : x_k > a_k\}$. Then $x_k \le a_k$,
$k < m$, so
$$f(x_1, x_2, \dots, x_n) \ge \frac{x_m}{1 + x_1 + x_2 + \dots + x_m} \\ \qquad > \frac{a_m}{1 + a_1 + a_2 + \dots + a_m} = f(a_1, a_2, \dots, a_n).$$
Consequently, the required minimum is $1 - 2^{-1/n}$.