4. Let $f(t)=\sqrt{1+t^{2}}-t$. Calculate the value of the expression

$$
f(x) f(y)+f(y) f(z)+f(z) f(x)
$$

if $x>0, y>0, z>0$ and $x y+y z+z x=1$.
Solution. We introduce trigonometric substitutions

$$
x=\operatorname{ctg} \alpha, y=\operatorname{ctg} \beta, \quad z=\operatorname{ctg} \gamma
$$

where $\alpha, \beta, \gamma$ are acute angles. From the condition of the problem, it follows that the equality holds

$$
\operatorname{ctg} \alpha \operatorname{ctg} \beta+\operatorname{ctg} \beta \operatorname{ctg} \gamma+\operatorname{ctg} \gamma \operatorname{ctg} \alpha=1
$$

which is equivalent to the equality

$$
\operatorname{tg} \alpha+\operatorname{tg} \beta+\operatorname{tg} \gamma=\operatorname{tg} \alpha \operatorname{tg} \beta \operatorname{tg} \gamma
$$

We will prove that there exists a triangle with angles $\alpha, \beta, \gamma$. Note that from the fact that the angles are acute, it follows that $(\operatorname{tg} \alpha+\operatorname{tg} \beta) \operatorname{tg} \gamma \neq 0$. From (1), we obtain the following equivalent equalities:

$$
\begin{aligned}
& \operatorname{tg} \alpha+\operatorname{tg} \beta+\operatorname{tg} \gamma=\operatorname{tg} \alpha \operatorname{tg} \beta \operatorname{tg} \gamma \\
& \operatorname{tg} \alpha+\operatorname{tg} \beta+\operatorname{tg} \gamma(1-\operatorname{tg} \alpha \operatorname{tg} \beta)=0 \\
& \frac{1}{\operatorname{tg} \gamma}+\frac{1-\operatorname{tg} \alpha \operatorname{tg} \beta}{\operatorname{tg} \alpha+\operatorname{tg} \beta}=0 \\
& \operatorname{ctg} \gamma+\operatorname{ctg}(\alpha+\beta)=0 \\
& \frac{\sin (\alpha+\beta+\gamma)}{\sin (\alpha+\beta) \sin \gamma}=0
\end{aligned}
$$

But, the angles $\alpha, \beta, \gamma$ are acute, so from (2) it follows that $\alpha+\beta+\gamma=\pi$.

With the substitution $x=\operatorname{ctg} \alpha$ and considering that $\sin \alpha>0$, the function takes the form

$$
f(x)=\sqrt{1+\operatorname{ctg}^{2} \alpha}-\operatorname{ctg} \alpha=\frac{1}{\sin \alpha}-\frac{\cos \alpha}{\sin \alpha}=\frac{1-\cos \alpha}{\sin \alpha}=\frac{2 \sin ^{2}(\alpha / 2)}{2 \sin (\alpha / 2) \cos (\alpha / 2)}=\operatorname{tg} \frac{\alpha}{2} .
$$

Similarly, we obtain $f(y)=\operatorname{tg} \frac{\beta}{2}$ and $f(z)=\operatorname{tg} \frac{\gamma}{2}$. Finally, for the value of the given expression, we get

$$
\begin{aligned}
f(x) f(y)+f(y) f(z)+f(z) f(x) & =\operatorname{tg} \frac{\alpha}{2} \operatorname{tg} \frac{\beta}{2}+\operatorname{tg} \frac{\beta}{2} \operatorname{tg} \frac{\gamma}{2}+\operatorname{tg} \frac{\gamma}{2} \operatorname{tg} \frac{\alpha}{2} \\
& =\operatorname{tg} \frac{\alpha}{2}\left(\operatorname{tg} \frac{\beta}{2}+\operatorname{tg} \frac{\gamma}{2}\right)+\operatorname{tg} \frac{\beta}{2} \operatorname{tg} \frac{\gamma}{2} \\
& =\operatorname{tg} \frac{\alpha}{2} \operatorname{tg} \frac{\beta+\gamma}{2}\left(1-\operatorname{tg} \frac{\beta}{2} \operatorname{tg} \frac{\gamma}{2}\right)+\operatorname{tg} \frac{\beta}{2} \operatorname{tg} \frac{\gamma}{2} \\
& =\operatorname{tg} \frac{\alpha}{2} \operatorname{ctg} \frac{\alpha}{2}\left(1-\operatorname{tg} \frac{\beta}{2} \operatorname{tg} \frac{\gamma}{2}\right)+\operatorname{tg} \frac{\beta}{2} \operatorname{tg} \frac{\gamma}{2}=1
\end{aligned}
$$

## 4th year