Exercise 14. 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 14 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 the $a_{i}$.

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 thus 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 therefore 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.

Grader's Comment: The exercise is very well solved! No student forgot to provide an example achieving equality.