Let's write the following expressions in a simpler form:

$$
\begin{aligned}
& \text { a) } \quad K=\sqrt{x+2 y \sqrt{x-y^{2}}}+\sqrt{x-2 y \sqrt{x-y^{2}}} \\
& \text { b) } \quad L=\frac{2 x y z}{\sqrt{\left(1+x^{2}\right)\left(1+y^{2}\right)\left(1+z^{2}\right)}}, \quad \text { if } x y+y z+z x=1
\end{aligned}
$$
I. Solution. a) The first expression is defined for real numbers if $y$ is any positive, negative number, or 0, and $x \geqq y^{2}$. Thus, the numbers under the large square roots are non-negative, since each is the square of a real number:

$$
x \pm 2 y \sqrt{x-y^{2}}=\left(x-y^{2}\right) \pm 2 y \sqrt{x-y^{2}}+y^{2}=\left(y \pm \sqrt{x-y^{2}}\right)^{2}
$$

Therefore, our expression can be written as:

$$
K=\left|y+\sqrt{x-y^{2}}\right|+\left|y-\sqrt{x-y^{2}}\right|
$$

We also notice that if we replace $y$ with $-y$, the value of $K$ remains unchanged, since only the values of its two terms are swapped; in other words: $K$ depends only on $|y|$. Now, depending on the magnitude of $|y|$ and $\sqrt{x-y^{2}}$, the value of $K$ can evolve in two ways:

I. If $\sqrt{x-y^{2}} \leqq|y|$, that is, if $x \leqq 2 y^{2}$ (and $x \geqq y^{2}$), then

$$
K=\left(|y|+\sqrt{x-y^{2}}\right)+\left(|y|-\sqrt{x-y^{2}}\right)=2 \cdot|y|
$$

If, however,

II. $|y| < \sqrt{x-y^{2}}$, that is, if $x > 2 y^{2}$, then

$$
K=\left(|y|+\sqrt{x-y^{2}}\right)+\left(\sqrt{x-y^{2}}-|y|\right)=2 \sqrt{x-y^{2}} .
$$

In other words, $K$ is equal to twice the larger of the expressions $|y|$ and $\sqrt{x-y^{2}}$.

b) The second expression is defined for all (real) $x, y, z$ numbers. Expanding the square of the denominator $N$,

$$
N^{2}=\left(1+x^{2}\right)\left(1+y^{2}\right)\left(1+z^{2}\right)=1+\left(x^{2}+y^{2}+z^{2}\right)+\left(x^{2} y^{2}+y^{2} z^{2}+z^{2} x^{2}\right)+x^{2} y^{2} z^{2}
$$

The two bracketed expressions can be transformed as follows based on the assumption:

$$
\begin{gathered}
x^{2}+y^{2}+z^{2}=(x+y+z)^{2}-2(x y+y z+z x)=(x+y+z)^{2}-2 \\
x^{2} y^{2}+y^{2} z^{2}+z^{2} x^{2}=(x y+y z+z x)^{2}-2\left(x y^{2} z+x y z^{2}+x^{2} y z\right)=1-2 x y z(x+y+z)
\end{gathered}
$$

Therefore,

$$
\begin{gathered}
N^{2}=(x+y+z)^{2}-2 x y z(x+y+z)+x^{2} y^{2} z^{2}=(x+y+z-x y z)^{2} \\
L=\frac{2 x y z}{|x+y+z-x y z|}
\end{gathered}
$$

Peter Fodor (Szeged, Ságvári E. Gymnasium, 1st year)

II. Solution for part b) of the second expression. We can also reach our goal by expressing one of the variables from the conditional equation and substituting it into $L$; for example,

$$
z=\frac{1-x y}{x+y}
$$

Here the denominator cannot be 0, that is, $y \neq -x$, because otherwise, according to the assumption, $-x^{2}=1$ would have to be true, which is impossible. According to (1),

$$
1+z^{2}=\frac{(x+y)^{2}+(1-x y)^{2}}{(x+y)^{2}}=\frac{x^{2}+y^{2}+1+x^{2} y^{2}}{(x+y)^{2}}=\frac{\left(1+x^{2}\right)\left(1+y^{2}\right)}{(x+y)^{2}}
$$

Thus,

$$
\begin{aligned}
& L=\frac{2 x y(1-x y)}{(x+y) \cdot \frac{\left(1+x^{2}\right)\left(1+y^{2}\right)}{|x+y|}}=\varepsilon \cdot \frac{2 x y(1-x y)}{\left(1+x^{2}\right)\left(1+y^{2}\right)} \\
& x+y \neq 0
\end{aligned}
$$

where $\varepsilon=1$ or -1 depending on whether $x+y$ is positive or negative.

Remarks. 1. It can be shown that in part b), $x+z$ and $y+z$ have the same sign as $x+y$.

2. The need to consider the sign also arises from the fact that the assumption holds for the number triplet $x, y, z$ as well as for the triplet $-x$, $-y, -z$, but the value of $L$ changes to its negative with this substitution.