Minimum of expression

Let $x_1, x_2, \dots, x_n$ be positive real numbers which sum to $1$. Find the minimum of:

$$\mathcal{A}=\max\left \{ \frac{x_1}{1+x_1}, \frac{x_2}{1+x_1+x_2}, \cdots, \frac{x_n}{1+x_1+x_2+\cdots+x_n} \right \}$$

Solution (G. Basdekis)

From the definition of $A$ we have that:

$$A\geq \frac{x_1}{1+x_1}, \; A\geq \frac{x_2}{1+x_1+x_2}, \; \dots, \; A \geq \frac{x_n}{1+x_1+x_2+\cdots+x_n}$$

Solving the first inequality we have that:

$$A\geq \frac{x_1}{1+x_1} \Leftrightarrow  x_1 \leq \frac{A}{1-A}$$

Solving the second , we get that:

$$A \geq \frac{x_2}{1+x_1+x_2}\geq \frac{x_2}{1+\frac{A}{1-A}+x_2} = \frac{x_2}{\frac{1}{1-A} +x_2} \Leftrightarrow x_2 \leq \frac{A}{\left ( 1-A \right )^2}$$

Continuing the procedure, we easily see that:

$$x_k \leq \frac{A}{\left ( 1-A \right )^k}$$

Summing up these $n$ inequalities we get:

$$\sum_{k=1}^{n}x_i \leq \sum_{k=1}^{n}\frac{A}{\left ( 1-A \right )^k}\Leftrightarrow  \sum_{k=1}^{n}\frac{A}{\left ( 1-A \right )^k} \geq 1$$

due to our initial assumption. Solving this last inequality we see that $\displaystyle A\geq  1- \frac{1}{\sqrt[n]{2}}$ and that is the minimum value of $A$ we are seeking.

The exercise was taken  from the Inequalities & Inequalities site of G. Bas. It can be found here