Problem 5. Let $r$ and $s$ be positive real numbers that satisfy the equation
$$
(r+s-r s)(r+s+r s)=r s .
$$

Find the minimum values of $r+s-r s$ and $r+s+r s$.
Solution. The given equation can be rewritten into
$$
(r+s)^{2}=r s(r s+1)
$$

Since $(r+s)^{2} \geq 4 r s$ for any $r, s \in \mathbb{R}$, it follows that $r s \geq 3$ for any $r, s>0$. Using this inequality, equation (1), and the assumption that $r$ and $s$ are positive, we have
$$
\begin{aligned}
r+s-r s=\sqrt{r s(r s+1)}-r s & =\frac{1}{\sqrt{1+\frac{1}{r s}}+1} \\
& \geq \frac{1}{\sqrt{1+\frac{1}{3}}+1}=-3+2 \sqrt{3} .
\end{aligned}
$$

Similarly, we also have
$$
r+s+r s \geq 3+2 \sqrt{3} \text {. }
$$

We show that these lower bounds can actually be attained. Observe that if $r=s=\sqrt{3}$, then
$$
r+s-r s=-3+2 \sqrt{3} \text { and } r+s+r s=3+2 \sqrt{3} .
$$

Therefore, the required minimum values of $r+s-r s$ and $r+s+r s$ are $-3+2 \sqrt{3}$ and $3+2 \sqrt{3}$, respectively.