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

$$
\begin{aligned}
& \left(\frac{p}{q}+\frac{q}{r}+\frac{r}{p}-1\right)(p+q+r)+\left(\frac{p}{q}+\frac{q}{r}-\frac{r}{p}+1\right)(p+q-r)+ \\
+ & \left(\frac{p}{q}-\frac{q}{r}+\frac{r}{p}+1\right)(p-q+r)+\left(-\frac{p}{q}+\frac{q}{r}+\frac{r}{p}+1\right)(-p+q+r)
\end{aligned}
$$
Multiplications offer opportunities for simplification. Indeed, in each of the four products of $K$, the two factors are equal in absolute value for their respective terms.

If we multiply everything by the first term of the second factors, we get $p$ times the sum of the first three terms' first factors minus the first factor of the fourth term. In this simplification, we get the same expression from the second, third, and fourth terms, each twice with a + sign and twice with a - sign, so after simplifying these partial products, $4 p^{2} / q$ remains. Similarly, in the sum of the terms to be multiplied by $q$, we need to subtract the third "large" parenthesis term from the sum of the other three large parenthesis terms, and in the case of $r$, the second large parenthesis expression is subtracted. Thus

$$
K=\frac{4 p}{q} \cdot p+\frac{4 q}{r} \cdot q+\frac{4 r}{p} \cdot r=4\left(\frac{p^{2}}{q}+\frac{q^{2}}{r}+\frac{r^{2}}{p}\right)
$$

Lidia Lukács (Püspökladány, F. Karacs II. o. t.)

Note. We can also reach the goal similarly if we only factor out the first factor from each product and form the appropriate sum or difference from the second factors.