Ionin Yu. I.

The sum of $n$ positive numbers $x_{1}, x_{2}, x_{3}, \ldots, x_{n}$ is 1.

Let $S$ be the maximum of the numbers $\frac{x_{1}}{1+x_{1}}, \frac{x_{2}}{1+x_{1}+x_{2}}, \ldots, \frac{x_{n}}{1+x_{1}+x_{2}+\ldots+x_{n}}$.

Find the smallest possible value of $S$. For which values of $x_{1}, x_{2}, \ldots, x_{n}$ is it achieved?
Suppose that the positive numbers $a_{1}, a_{2}, \ldots, a_{n}$ are such that $a_{1}+a_{2}+\ldots+a_{n}=1$ and $\frac{a_{1}}{1+a_{1}}=\frac{a_{2}}{1+a_{1}+a_{2}}=\ldots=\frac{a_{n}}{1+a_{1}+a_{2}+\ldots+a_{n}}$. (We will see below that such a set of $a_{1}, a_{2}, \ldots, a_{n}$ indeed exists.) We will show that for $x_{1}=a_{1}, x_{2}=a_{2}, \ldots, x_{n}=a_{n}$, the value of $S$ attains its minimum possible value.

Let $b_{1}, b_{2}, \ldots, b_{n}$ be another set of positive numbers whose sum is 1. Clearly, $b_{k}>a_{k}$ for some $k$. Let $k$ be the smallest index with this property, that is, $b_{1} \leq a_{1}, b_{2} \leq a_{2}, \ldots, b_{k-1} \leq a_{k-1}, b_{k}>a_{k}$, then $\frac{b_{k}}{1+b_{1}+\ldots+b_{k}}=\left(\frac{1+b_{1}+\ldots+b_{k-1}}{b_{k}}+1\right)^{-1}>\left(\frac{1+a_{1}+\ldots+a_{k-1}}{a_{k}}+1\right)^{-1}=$ $\frac{a_{k}}{1+a_{1}+\ldots+a_{k}}=\frac{a_{1}}{1+a_{1}}$

Thus, the largest of the numbers $\frac{b_{1}}{\frac{1}{a_{2}+b_{1}}}, \frac{b_{2}}{1+b_{1}+b_{2}}, \ldots, \frac{b_{n}}{a_{n}}, \frac{a_{1}}{1+b_{1}+b_{2}+\ldots+b_{n}}$ is greater than $\frac{a_{1}}{1+a_{1}}$ (the largest of the numbers $\left.\frac{a_{1}}{1+a_{1}}, \frac{a_{2}}{1+a_{1}+a_{2}}, \ldots, \frac{a_{n}}{1+a_{1}+a_{2}+\ldots+a_{n}}\right)$.

It remains to solve the system of equations in positive numbers $x_{1}+x_{2}+\ldots+x_{n}=1, \frac{x_{1}}{1+x_{1}}=\frac{x_{2}}{1+x_{1}+x_{2}}=\ldots=\frac{x_{n}}{1+x_{1}+x_{2}+\ldots+x_{n}}$ The equation $\frac{x_{1}}{1+x_{1}}=\frac{x_{k}}{1+x_{1}+\ldots+x_{k}}$ (for positive $\left.x_{1}, \ldots, x_{k}\right)$ transforms to $x_{k}=\left(1+x_{1}+\right.$ $\left.\ldots+x_{k-1}\right) x_{1}$, so the given system is equivalent to the system

$x_{2}=\left(1+x_{1}\right) x_{1}$,

$x_{3}=\left(1+x_{1}+x_{2}\right) x_{1}$,

$$
x_{n}=\left(1+x_{1}+x_{2}+\ldots+x_{n-1}\right) x_{1}
$$

$$
x_{1}+x_{2}+\ldots+x_{n}=1
$$

which in turn is equivalent to the system

$x_{2}=\left(1+x_{1}\right) x_{1}$,

$x_{3}=\left(1+x_{1}\right)^{2} x_{1}$,

$\cdots$

$x_{n}=\left(1+x_{1}\right)^{n-1} x_{1}$

$x_{1}+x_{2}+\ldots+x_{n}=1$

that is, the numbers $x_{1}, x_{2}, \ldots, x_{n}$ form a geometric progression with common ratio $1+x_{1}$ and sum $\left(1+x_{1}\right)^{n}-$

1. Now the solution to the system is easily found:

$a_{1}=\sqrt[n]{2}-1, a_{2}=\sqrt[n]{2}(\sqrt[n]{2}-1), \ldots, a_{n}=\sqrt[n]{2^{n-1}}(\sqrt[n]{2}-1)$

## Answer

$S=1-\frac{1}{\sqrt[n]{2}}$ for $x_{k}=\sqrt[n]{2^{k-1}}(\sqrt[n]{2}-1)(k=1, \ldots, n)$