1. determine all positive real solutions of the following system of equations:

$$
\begin{aligned}
& a=\max \left\{\frac{1}{b}, \frac{1}{c}\right\} \\
& b=\max \left\{\frac{1}{c}, \frac{1}{d}\right\} \\
& c=\max \left\{\frac{1}{d}, \frac{1}{e}\right\} \\
& d=\max \left\{\frac{1}{e}, \frac{1}{f}\right\} \\
& e=\max \left\{\frac{1}{f}, \frac{1}{a}\right\} \\
& f=\max \left\{\frac{1}{a}, \frac{1}{b}\right\}
\end{aligned}
$$
## 1st solution

Because of the cyclic symmetry of the system, we can assume that $a$ is maximal among all six numbers. From this follows directly

$$
f=\max \left\{\frac{1}{a}, \frac{1}{b}\right\}=\frac{1}{b}, \quad e=\max \left\{\frac{1}{f}, \frac{1}{a}\right\}=\frac{1}{f}=b, \quad d=\max \left\{\frac{1}{e}, \frac{1}{f}\right\}=\max \left\{b, \frac{1}{b}\right\}
$$

We now distinguish between two cases:

(1) Let $b \geq 1$. Then $d=b$ and you get

$$
c=\max \left\{\frac{1}{d}, \frac{1}{e}\right\}=\frac{1}{b}, \quad a=\max \left\{\frac{1}{b}, \frac{1}{c}\right\}=\max \left\{b, \frac{1}{b}\right\}=b
$$

(2) Let $b \leq 1$. Then $d=\frac{1}{b}$ and you get

$$
c=\max \left\{\frac{1}{d}, \frac{1}{e}\right\}=\max \left\{b, \frac{1}{b}\right\}=\frac{1}{b}, \quad a=\max \left\{\frac{1}{b}, \frac{1}{c}\right\}=\frac{1}{b}
$$

A comparison of the results shows that both cases give the same solutions, except for a cyclic shift. These are the cyclic permutations of

$$
(a, b, c, d, e, f)=\left(x, x, \frac{1}{x}, x, x, \frac{1}{x}\right), \quad x \geq 1
$$

To show that these are indeed all solutions, it is again sufficient to confirm the equations for $a$ and $c$ because of the symmetry. In fact, because of $x \geq 1$

$$
x=a=\max \left\{\frac{1}{b}, \frac{1}{c}\right\}=\max \left\{\frac{1}{x}, x\right\}=x, \quad \frac{1}{x}=c=\max \left\{\frac{1}{d}, \frac{1}{e}\right\}=\max \left\{\frac{1}{x}, \frac{1}{x}\right\}=\frac{1}{x}
$$