Exercise 7. Let $n \geqslant 2$ be a fixed integer, and let $a_{1}, \ldots, a_{n}$ be strictly positive real numbers such that $a_{1}+ \ldots + a_{n} = 2^{n} - 1$. Determine the smallest value that can be taken by:

$$
\frac{a_{1}}{1}+\frac{a_{2}}{a_{1}+1}+\frac{a_{3}}{1+a_{1}+a_{2}}+\ldots+\frac{a_{n}}{1+a_{1}+a_{2}+\ldots+a_{n-1}}
$$
Solution to Exercise 7 The expression in question involves the sums of the first terms of the sequence $\left(a_{i}\right)$, so that each fraction quickly involves a large number of terms of the sequence. One way to handle these sums is to change the variable and consider the sequence of sums of $a_{i}$.

Let $a_{0}=1$ and $b_{k}=a_{0}+a_{1}+\ldots+a_{k}$ for $0 \leqslant k \leqslant n$. We have

$$
\frac{a_{1}}{1}+\frac{a_{2}}{a_{1}+1}+\frac{a_{3}}{1+a_{1}+a_{2}}+\ldots+\frac{a_{n}}{1+a_{1}+a_{2}+\ldots+a_{n-1}}=\frac{b_{1}-b_{0}}{b_{0}}+\ldots+\frac{b_{n}-b_{n-1}}{b_{n-1}}
$$

Here, we see that each fraction involves at most two terms of the sequence $\left(b_{k}\right)$, making them much more manageable.

We now have:

$$
\begin{aligned}
& \frac{b_{1}-b_{0}}{b_{0}}+\ldots+\frac{b_{n}-b_{n-1}}{b_{n-1}} \\
= & \frac{b_{1}}{b_{0}}+\ldots+\frac{b_{n}}{b_{n-1}}-n \\
\geqslant & n \sqrt[n]{\frac{b_{1}}{b_{0}} \cdot \ldots \cdot \frac{b_{n}}{b_{n-1}}}-n \\
= & n\left(\sqrt[n]{2^{n}}-1\right) \\
= & n
\end{aligned}
$$

We have equality if and only if each of the $\frac{b_{i}}{b_{i-1}}$ for $1 \leqslant i \leqslant n$ is equal, in particular equal to the geometric mean of these elements, which is 2. We therefore deduce that $b_{i}=2 b_{i-1}$ for all integers $i$ such that $1 \leqslant i \leqslant n$.

In particular, by iterating this result, we obtain that for all $i$ between 0 and $n, b_{i}=2^{i} b_{0}=2^{i}$, and thus for all integers $i$ between 1 and $n, a_{i}=b_{i}-b_{i-1}=2^{i}-2^{i-1}=2^{i-1}$.

In particular, if $a_{i}=2^{i-1}$ for $i$ between 1 and $n$, we get $b_{i}=1+1+2+\cdots+2^{i-1}=1+2^{i}-1=2^{i}$ for all integers $i$ between 1 and $n$, and for $i=0$. We then have that $\frac{b_{i}}{b_{i-1}}=2$ for all integers $i$ between 1 and $n$, hence the equality.

Comment from the graders: The exercise is well solved by the students, with the detail that some students forget to verify that the minimum they find can indeed be achieved by providing an example of equality. This omission is systematically penalized in competitions.