Let $n$ be a positive integer greater than two. What is the largest value of $h$ and the smallest value of $H$ for which

$$
h<\frac{a_{1}}{a_{1}+a_{2}}+\frac{a_{2}}{a_{2}+a_{3}}+\ldots+\frac{a_{n}}{a_{n}+a_{1}}<H
$$

is true for any positive numbers $a_{1}, a_{2}, \ldots, a_{n}$?
I. solution. Let's look at the sum of two consecutive terms:

$$
\frac{a_{i}}{a_{i}+a_{i+1}}+\frac{a_{i+1}}{a_{i+1}+a_{i+2}}
$$

where the term after the last one refers to the first. Accordingly, $a_{n+1}, a_{n+2}$ represent $a_{1}$ and $a_{2}$, respectively. The sum of the two terms is at least 1 if $a_{i+2} \leq a_{i}$, and at most 1 if $a_{i+2} \geq a_{i}$.

Choose $i$ such that $a_{i+2}$ is the smallest value among those that occur. Then the sum of the selected two terms is at least 1. The other terms are all positive, and by condition, there is at least one more term, so the sum is always greater than 1. If, however, $i$ is chosen such that $a_{i+2}$ is the largest value among those that occur, then the sum of the selected two terms is at most 1; since the other $n-2$ terms are each less than 1, we get that the sum is always less than

$$
1+n-2=n-1
$$

We will show that the sum can get arbitrarily close to both bounds if the sequence $a_{i}$ is chosen appropriately. For the geometric sequence with positive ratio $q$:

$$
a_{i}=q^{i-1}, \quad i=1,2, \ldots, n
$$

the sum $S_{q}$ is:

$$
S_{q}=\frac{n-1}{1+q}+\frac{q^{n-1}}{q^{n-1}+1}
$$

Let's estimate this from above. For any (small) positive number $p$,

$$
S_{q}\frac{n-1}{p}
$$

On the other hand, estimating $S_{q}$ from below,

$$
S_{q}>\frac{n-1}{1+q}=n-1-\frac{q(n-1)}{1+q}>(n-1)-(n-1) q>n-1-p, \quad \text { if } \quad q1
$$

Now suppose that the sums formed from the $n$ positive numbers are always greater than 1, where $n \geq 3$. Then it is enough to show that the sum of $a_{1}, a_{2}, \ldots, a_{n+1}$ is greater than that formed from the first $n$ numbers. But

$$
\begin{gathered}
T_{n+1}-T_{n}=\frac{a_{n}}{a_{n}+a_{n+1}}+\frac{a_{n+1}}{a_{n+1}+a_{1}}-\frac{a_{n}}{a_{n}+a_{1}}= \\
=\frac{a_{n}}{a_{n}+a_{n+1}}+\frac{a_{n+1}}{a_{n+1}+a_{1}}+\frac{a_{1}}{a_{1}+a_{n}}-1>0
\end{gathered}
$$

because the sum of the three fractions is the sum formed from the numbers $a_{n}, a_{n+1}$, and $a_{1}$, and we have already shown that it is always greater than 1. Thus, using the inductive hypothesis,

$$
T_{n+1}>T_{n}>1
$$

and this is what we needed to prove.

Remarks. 1. The last proof could have been significantly simplified by the observation that $T_{n+1}$ does not change if the numbers are cyclically permuted. Thus, we can choose the order such that $a_{n+1}$ is the smallest value that occurs. Then the first term in the difference $T_{n+1}-T_{n}$ is not less than the subtrahend, so the difference is positive. Moreover, the sum in question can be formed from two numbers and its value is 1, so for any $n$ greater than 2, it is already greater than 1. This would avoid all calculations.

However, the above solution shows that the inductive proof leads to the goal without any further catches. In this case, it would not have been worth avoiding the calculation for the case of 3 terms, as we could also use it in the second part of the solution.

2. The sum in question actually depends only on the ratios $a_{i+1} / a_{i}$. Let these be denoted by $b_{i}$ for $i=1,2, \ldots, n-1$ and let $b_{n}=a_{n} / a_{1}$. Then the sum in question takes the form

$$
\frac{1}{1+b_{1}}+\frac{1}{1+b_{2}}+\ldots+\frac{1}{1+b_{n}}
$$

and here

$$
b_{1} b_{2} \ldots b_{n}=1
$$

Combine the last two terms:

$$
\frac{1}{1+b_{n-1}}+\frac{1}{1+b_{n}}>\frac{1+b_{n-1}+b_{n}}{1+b_{n-1} b_{n}+b_{n-1}+b_{n}}>\frac{1}{1+b_{n-1} b_{1}}
$$

based on the remark used in the second solution. Inserting this into the sum, we get the $(n-1)$-term sum formed from the numbers $b_{1}, \ldots, b_{n-2}, b_{n-1} b_{n}$, and the product of the numbers is still 1. Based on this, we can obtain another inductive proof.