6. Let $[x]$ denote the greatest integer not exceeding $x$. If $p, q, r$ are positive, find the minimum value of:
$$
\left[\frac{p+q}{r}\right]+\left[\frac{q+r}{p}\right]+\left[\frac{r+p}{q}\right]
$$
6. 4
6. Note that for positive real number $x$, we have $[x]>x-1$ and $x+\frac{1}{x} \geq 2$. Hence
$$
\begin{aligned}
{\left[\frac{p+q}{r}\right]+\left[\frac{q+r}{p}\right]+\left[\frac{r+p}{q}\right] } & >\left(\frac{p+q}{r}-1\right)+\left(\frac{q+r}{p}-1\right)+\left(\frac{r+p}{q}-1\right) \\
& =\left(\frac{p}{q}+\frac{q}{p}\right)+\left(\frac{q}{r}+\frac{r}{q}\right)+\left(\frac{r}{p}+\frac{p}{r}\right)-3 \\
& \geq 2+2+2-3 \\
& =3
\end{aligned}
$$

Since $\left[\frac{p+q}{r}\right]+\left[\frac{q+r}{p}\right]+\left[\frac{r+p}{q}\right]$ is an integer, it must be at least 4. Such minimum is attainable, for example when $p=6, q=8$ and $r=9$.

Note: The text provided is already in English, so no translation was necessary.