XXXI Brazilian Math Olympiad

The answer is all real numbers in the interval $]1,\lfloor n/2\rfloor [$. For simplicity, let $$E=\frac{x_1}{x_n+x_1+x_2}+\frac{x_2}{x_1+x_2+x_3}+\frac{x_3}{x_2+x_3+x_4}+\cdots +\frac{x_n}{x_{n-1}+x_n+x_1}.$$ Let's prove first the lower bound. Let $S=x_1+x_2+\cdots +x_n$. First notice that $$E>\frac{x_1}{S}+\frac{x_2}{S}+\frac{x_3}{S}+\cdots +\frac{x_n}{S}=\frac{x_1+x_2+\cdots +x_n}{S}=1.$$ By making, say, $x_i={ϵ}^{i-1}$, we obtain $\frac{x_i}{x_{i-1}+x_i+x_{i+1}}=\frac{ϵ}{1+ϵ+{ϵ}^2}$ for $1<i<n$, $\frac{x_1}{x_n+x_1+x_2}=\frac{1}{{ϵ}^{n-1}+1+ϵ}$ and $\frac{x_n}{x_{n-1}+x_n+x_1}=\frac{{ϵ}^{n-1}}{{ϵ}^{n-2}+{ϵ}^{n-1}+1}$. Thus, by making $ϵ$ very small we obtain $E$ arbitrarily close to 1.

Now, for the upper bound, notice that, for $n$ even, $$E<\frac{x_1}{x_1+x_2}+\frac{x_2}{x_1+x_2}+\frac{x_3}{x_3+x_4}+\frac{x_4}{x_3+x_4}+\cdots +\frac{x_n}{x_{n-1}+x_n}=\frac{n}{2}=\lfloor \frac{n}{2}\rfloor$$ For $n$ odd, suppose without loss of generality that the minimum of the denominators $x_n+x_1+x_2,x_1+x_2+x_3,\dots ,x_{n-1}+x_n+x_1$ is $x_1+x_2+x_3$. Thus $$\frac{x_1}{x_n+x_1+x_2}+\frac{x_2}{x_1+x_2+x_3}+\frac{x_3}{x_2+x_3+x_4}\le \frac{x_1}{x_1+x_2+x_3}+\frac{x_2}{x_1+x_2+x_3}+\frac{x_3}{x_1+x_2+x_3}=1$$ This implies $$\begin{array}{rl}E<1& +\frac{x_4}{x_4+x_5}+\frac{x_5}{x_4+x_5}+\frac{x_6}{x_6+x_7}+\frac{x_7}{x_6+x_7}+\cdots +\frac{x_{n-1}}{x_{n-1}+x_n}+\frac{x_n}{x_{n-1}+x_n}\\ =1& +\frac{n-3}{2}=\lfloor \frac{n}{2}\rfloor \end{array}$$ Now, to attain the upper bound, choose $x_{2k}=ϵ$ and $x_{2k-1}=1$, $k=1,2,\dots ,\lfloor n/2\rfloor$. For $n$ even, we have the summands $\frac{x_{2k-1}}{x_{2k-2}+x_{2k-1}+x_{2k}}=\frac{1}{1+2ϵ}$ and $\frac{x_{2k}}{x_{2k-1}+x_{2k}+x_{2k+1}}=\frac{ϵ}{2+ϵ}$ and $E$ gets arbitrarily close to $n/2$ as $ϵ$ gets small. For $n$ odd, notice that $x_1=x_n=ϵ$, so $\frac{x_{2k-1}}{x_{2k-2}+x_{2k-1}+x_{2k}}=\frac{1}{1+2ϵ}$ and $\frac{x_{2k}}{x_{2k-1}+x_{2k}+x_{2k+1}}=\frac{ϵ}{2+ϵ}$ and $E$ gets arbitrarily close to $\lfloor n/2\rfloor$ as $ϵ$ gets small.