, 2008

$$f(x) = x \quad \text{and} \quad f(x) = \frac{1}{x}.$$
It is easy to check that these two functions satisfy the conditions of the problem. We now show that they are the only functions satisfying the conditions of the problem.
Setting $p = q = r = s = 1$ in $(*)$ gives
$$f(1) = \frac{(f(1))^2 + (f(1))^2}{f(1) + f(1)} = \frac{1^2 + 1^2}{1^1 + 1^2} = 1.$$
For positive real numbers $x$, setting $(p, q, r, s) = (1, x, \sqrt{x}, \sqrt{x})$ in $(*)$ yields
$$\frac{1 + (f(x))^2}{2f(x)} = \frac{(f(1))^2 + (f(x))^2}{2f(x)} = \frac{1 + x^2}{2x} \quad \text{or} \quad 2x + 2x(f(x))^2 = 2f(x) + 2x^2f(x).$$
We deduce that
$$0 = x(f(x))^2 - x^2 f(x) + x - f(x) = (x - f(x))(1 - x f(x)).$$
It follows that for positive real numbers $x$,
$$\text{either } f(x) = x \quad \text{or} \quad f(x) = \frac{1}{x}.$$
Let us assume that $f(x) \neq x$ and $f(x) \neq \frac{1}{x}$. Then there are positive real numbers $a$ and $b$ such that $f(a) \neq a$ and $f(b) \neq \frac{1}{b}$. We deduce that $f(a) = \frac{1}{a}$ and $f(b) = b$. Setting $(p, q, r, s) = (a, b, \sqrt{ab}, \sqrt{ab})$ in $(*)$ gives
$$\frac{\frac{1}{a^2} + b^2}{2f(ab)} = \frac{(f(a))^2 + (f(b))^2}{2f(ab)} = \frac{a^2 + b^2}{2ab}$$
or
$$\frac{f(ab)}{ab} = \frac{1 + a^2 b^2}{a^2(a^2 + b^2)} = \frac{1 + a^2 b^2}{a^4 + a^2 b^2}. \qquad (\dagger)$$
We know that either $f(ab) = ab$ or $f(ab) = \frac{1}{ab}$. If $f(ab) = ab$, then $(\dagger)$ implies that $a^4 = 1$ or $a = 1$, but then $f(a) = f(1) = 1 = a$, violating our assumption of $f(a) \neq a$. If $f(ab) = \frac{1}{ab}$, then $(\dagger)$ implies that $b^2 + a^2 b^4 = a^2 + b^2$ or $b = 1$, but then $f(b) = f(1) = 1 = \frac{1}{b}$, violating our assumption of $f(b) \neq \frac{1}{b}$. We reach contradictions in both cases. Thus, our assumption was wrong and no such $a$ and $b$ exist. In other words, $f(x) = x$ and $f(x) = \frac{1}{x}$ are the only solutions of the problem.