1. Let $n \geq 3$ be a fixed natural number. Determine, for all positive real numbers $a_{1}, a_{2}, \ldots, a_{n}$, the minimum value of the sum $S=\left[\frac{a_{1}+a_{2}}{a_{3}}\right]+\left[\frac{a_{2}+a_{3}}{a_{4}}\right]+\ldots+\left[\frac{a_{n}+a_{1}}{a_{2}}\right]$.

Marius Marchitan, Suceava
## Solution.

$\overline{\text { Since }[x]}>x-1$, for any real number $x$, we deduce that:

$S>\frac{a_{1}+a_{2}}{a_{3}}-1+\frac{a_{2}+a_{3}}{a_{4}}-1+\ldots+\frac{a_{n}+a_{1}}{a_{2}}-1=\left(\frac{a_{1}}{a_{3}}+\frac{a_{2}}{a_{4}}+\ldots+\frac{a_{n}}{a_{2}}\right)+\left(\frac{a_{2}}{a_{3}}+\frac{a_{3}}{a_{4}}+\ldots+\frac{a_{1}}{a_{2}}\right)-n \geq$

$\geq n \cdot \sqrt[n]{\frac{a_{1}}{a_{3}} \cdot \frac{a_{2}}{a_{4}} \cdot \ldots \cdot \frac{a_{n}}{a_{2}}}+n \cdot \sqrt[n]{\frac{a_{2}}{a_{3}} \cdot \frac{a_{3}}{a_{4}} \cdot \ldots \cdot \frac{a_{1}}{a_{2}}}-n=n+n-n=n$

It follows that $S>n$ and since $S$ is a natural number, we deduce that $S \geq n+1$. To show that the minimum value of the sum $S$ is $n+1$, it remains to find an example of positive real numbers $a_{1}, a_{2}, \ldots, a_{n}$ for which $S=n+1$.

Let $a_{1}=n^{2}-(2 n-3)$, and the next $n-1$ numbers are consecutive:

$a_{2}=n^{2}-(n-2), a_{3}=n^{2}-(n-3), \ldots, a_{n-1}=n^{2}-1, a_{n}=n^{2}$. We then have:

$\left[\frac{a_{1}+a_{2}}{a_{3}}\right]=\left[\frac{n^{2}-(2 n-3)+n^{2}-(n-2)}{n^{2}-(n-3)}\right]=\left[\frac{2 n^{2}-3 n+5}{n^{2}-n+3}\right]=\left[1+\frac{n^{2}-2 n+2}{n^{2}-n+3}\right]=1$.

For any $k=\overline{2, n-2}$ :

$\left[\frac{a_{k}+a_{k+1}}{a_{k+2}}\right]=\left[\frac{n^{2}-(n-k)+n^{2}-(n-k-1)}{n^{2}-(n-k-2)}\right]=\left[\frac{2 n^{2}-2 n+2 k+1}{n^{2}-n+k+2}\right]=\left[1+\frac{n^{2}-n+k-1}{n^{2}-n+k+2}\right]=1$.

For the penultimate term of the sum, we find:

$\left[\frac{a_{n-1}+a_{n}}{a_{1}}\right]=\left[\frac{n^{2}-1+n^{2}}{n^{2}-(2 n-3)}\right]=\left[\frac{2 n^{2}-1}{n^{2}-2 n+3}\right]=\left[2+\frac{4 n-7}{n^{2}-2 n+3}\right]=2$,

since $4 n-7 \geq 0$ and $4 n-7x-1, \forall x \in \mathbb{R}$ | $1 \text{p}$ |
| :--- | :--- |
| Deduce $S>\frac{a_{1}+a_{2}}{a_{3}}+\frac{a_{2}+a_{3}}{a_{4}}+\ldots+\frac{a_{n}+a_{1}}{a_{2}}-n$ | $1 \text{p}$ |
| Obtain $S>n$ | $2 \text{p}$ |
| Deduce $S \geq n+1$ | $1 \text{p}$ |
| Justify with an example that the minimum value of $S$ is $n+1$ | $2 \text{p}$ |