3. Let $f:(0,+\infty) \rightarrow \mathbb{R}$ be a function that satisfies the following conditions:
a) $f$ is strictly increasing;

b) $f(x) > -\frac{1}{x}$, for $x > 0$;

c) $f(x) f\left(f(x) + \frac{1}{x}\right) = 1$, for $x > 0$.

Calculate $f(1)$.
Solution. Let $x>0$ and $k=f(x)+\frac{1}{x}$. Then from condition b) it follows that $k>0$, so from condition c) it follows that

$$
f(k) f\left(f(k)+\frac{1}{k}\right)=1
$$

However, since $x>0$, from condition c) we get

$$
f(x) f(k)=f(x) f\left(f(x)+\frac{1}{x}\right)=1
$$

From the last two equalities, it follows that

$$
f(x)=f\left(f(k)+\frac{1}{k}\right)=f\left(\frac{1}{f(x)}+\frac{1}{f(x)+\frac{1}{x}}\right)
$$

Furthermore, the function $f$ is strictly increasing, so it is an injection, and from the last equality, it follows that $x=\frac{1}{f(x)}+\frac{1}{f(x)+\frac{1}{x}}$. We solve the last equation for $f(x)$ and get $f(x)=\frac{1 \pm \sqrt{5}}{2 x}$. It is easily verified that only the function $f(x)=\frac{1-\sqrt{5}}{2 x}$ satisfies the conditions of the problem. Therefore, $f(1)=\frac{1-\sqrt{5}}{2}$.