Romania — NMO Selection Tests for the Balkan and International Mathematical Olympiads

Only the first inequality will be proved; the second is dealt with similarly. Suppose, if possible, that each of the $n+1$ positive real numbers $x_1, 1/x_1 + x_2, \dots, 1/x_{n-1} + x_n, 1/x_n$ is greater than $2 \cos \alpha$, where $\alpha = \pi/(n+2)$, and show recursively that $x_k > \sin((k+1)\alpha) / \sin(k\alpha)$, $k = 1, \dots, n$; notice that $\sin((k+1)\alpha) / \sin(k\alpha) > 0$, $k = 1, \dots, n$. By assumption, $x_1 > 2 \cos \alpha = \sin(2\alpha) / \sin \alpha$. For the induction step, $x_{k+1} > 2 \cos \alpha - 1/x_k > 2 \cos \alpha - \sin(k\alpha) / \sin((k+1)\alpha) = \sin((k+2)\alpha) / \sin((k+1)\alpha)$. Consequently, $x_n > \sin((n+1)\alpha) / \sin(n\alpha) = 1/(2 \cos \alpha)$, in contradiction with $1/x_n > 2 \cos \alpha$; here, the last equality holds, since $\alpha = \pi/(n+2)$.

Remark. Notice that the $x_k = \sin((k+1)\alpha) / \sin(k\alpha)$, $k = 1, \dots, n$, where $\alpha = \pi/(n+2)$, satisfy
$$x_1 = 1/x_1 + x_2 = \dots = 1/x_{n-1} + x_n = 1/x_n = 2 \cos \alpha,$$
to conclude, by the preceding, that
$$\begin{aligned} \max_{x_1>0, \dots, x_n>0} \min (x_1, 1/x_1 + x_2, \dots, 1/x_{n-1} + x_n, 1/x_n) &= \\ \min_{x_1>0, \dots, x_n>0} \max (x_1, 1/x_1 + x_2, \dots, 1/x_{n-1} + x_n, 1/x_n) &= 2 \cos \alpha. \end{aligned}$$

We shall actually prove that
$$\begin{aligned} \max_{x_1>0, \dots, x_n>0} \min (x_1, 1/x_1 + x_2, \dots, 1/x_{n-1} + x_n, 1/x_n) &= \\ \min_{x_1>0, \dots, x_n>0} \max (x_1, 1/x_1 + x_2, \dots, 1/x_{n-1} + x_n, 1/x_n) &= 2 \cos \left(\frac{\pi}{n+2}\right). \end{aligned}$$
To this end, let $U$ denote the set of all $n$-tuples of positive real numbers, and, for $\mathbf{x} = (x_1, \dots, x_n)$ in $U$, let
$$m(\mathbf{x}) = \min (x_1, 1/x_1 + x_2, \dots, 1/x_{n-1} + x_n, 1/x_n)$$
and
$$M(\mathbf{x}) = \max (x_1, 1/x_1 + x_2, \dots, 1/x_{n-1} + x_n, 1/x_n).$$
The first step consists in assuming that $m(\mathbf{a}) = M(\mathbf{a})$ for some $\mathbf{a} = (a_1, \dots, a_n)$ in $U$ and showing that $m(\mathbf{x}) \le m(\mathbf{a}) = M(\mathbf{a}) \le M(\mathbf{x})$ for all $\mathbf{x}$ in $U$. Clearly, the condition $m(\mathbf{a}) = M(\mathbf{a})$ is equivalent to
$$a_1 = 1/a_1 + a_2 = \cdots = 1/a_{n-1} + a_n = 1/a_n. \tag{1}$$
Suppose, if possible, that $m(\mathbf{x}) > m(\mathbf{a})$ for some $\mathbf{x} = (x_1, \dots, x_n)$ in $U$. Then $x_1 \ge m(\mathbf{x}) > m(\mathbf{a}) = a_1$; $1/x_k + x_{k+1} \ge m(\mathbf{x}) > m(\mathbf{a}) = 1/a_k + a_{k+1}$, $k = 1, \dots, n-1$; and $1/x_n \ge m(\mathbf{x}) > m(\mathbf{a}) = 1/a_n$. The first $n$ inequalities imply recursively that $x_k > a_k$, $k = 1, \dots, n$; in particular, $x_n > a_n$, in contradiction with $1/x_n > 1/a_n$. Consequently, $m(\mathbf{x}) \le m(\mathbf{a})$ for all $\mathbf{x}$ in $U$. Similarly, $M(\mathbf{x}) \ge M(\mathbf{a})$ for all $\mathbf{x}$ in $U$.

To show the existence of an $\mathbf{a}$ in $U$ satisfying (1), let $a$ denote the common value in (1) and notice that $a_k = b_k / b_{k-1}$, $k = 1, \dots, n$, where the $b_k$ are defined by
$$b_0 = 1, \quad b_1 = a, \quad \text{and} \quad b_k = a b_{k-1} - b_{k-2}, \quad k \ge 2. \tag{2}$$
Since $1/a_n = a$, it follows that $b_{n-1} = a b_n$ which is equivalent to $b_{n+1} = 0$. Notice further that $a < 2$. Otherwise, $a_1 = a \ge 2$ and $a_k = a - 1/a_{k-1}$, $k = 2, \dots, n$, would recursively imply that $a_k \ge 1 + 1/k$, $k = 1, \dots, n$; in particular, $a_n \ge 1 + 1/n$, in contradiction with $1/a_n = a \ge 2$. We may therefore write $a = 2 \cos \alpha$, for some $\alpha$ in the open interval $(0, \pi/2)$, to deduce that the unique solution of (2) is $b_k = \sin((k+1)\alpha)/\sin \alpha$. Since $b_1, \dots, b_n$ are all positive, the condition $b_{n+1} = 0$ yields $\alpha = \pi/(n+2)$ and the conclusion follows.