Let $r$ and $s$ be positive real numbers such that $(r+s-rs)(r+s+rs)=rs$. Find the minimum value of $r+s-rs$ and $r+s+rs$
Given the equation \((r+s-rs)(r+s+rs)=rs\), we need to find the minimum values of \(r+s-rs\) and \(r+s+rs\).

1. Let \(r+s-rs = a\) and \(r+s+rs = b\). Then we have:
   \[
   r+s = \frac{a+b}{2} \quad \text{and} \quad rs = \frac{b-a}{2}
   \]

2. Since \(r\) and \(s\) are positive real numbers, we must have \(r+s > 0\) and \(rs > 0\). The discriminant of the quadratic equation \(t^2 - (r+s)t + rs = 0\) must be non-negative:
   \[
   (r+s)^2 - 4rs \geq 0
   \]
   Substituting \(r+s = \frac{a+b}{2}\) and \(rs = \frac{b-a}{2}\), we get:
   \[
   \left(\frac{a+b}{2}\right)^2 - 4 \cdot \frac{b-a}{2} \geq 0
   \]
   Simplifying, we obtain:
   \[
   \frac{(a+b)^2}{4} - 2(b-a) \geq 0
   \]
   Multiplying through by 4:
   \[
   (a+b)^2 \geq 8(b-a)
   \]

3. The given condition is \(ab = \frac{b-a}{2}\). Since \(ab > 0\), we have:
   \[
   (a+b)^2 \geq 8 \cdot 2ab \implies (a-b)^2 + 4ab \geq 16ab \implies (a-b)^2 \geq 12ab
   \]

4. From \(ab = \frac{b-a}{2}\), we can express \(b\) in terms of \(a\):
   \[
   b = \frac{a}{1-2a}
   \]
   Substituting this into the inequality \((a-b)^2 \geq 12ab\), we get:
   \[
   \left(\frac{a}{1-2a} - a\right)^2 \geq 12a \cdot \frac{a}{1-2a}
   \]
   Simplifying, we obtain:
   \[
   \left(\frac{a - a(1-2a)}{1-2a}\right)^2 \geq \frac{12a^2}{1-2a}
   \]
   \[
   \left(\frac{a - a + 2a^2}{1-2a}\right)^2 \geq \frac{12a^2}{1-2a}
   \]
   \[
   \left(\frac{2a^2}{1-2a}\right)^2 \geq \frac{12a^2}{1-2a}
   \]
   \[
   \frac{4a^4}{(1-2a)^2} \geq \frac{12a^2}{1-2a}
   \]
   Multiplying both sides by \((1-2a)^2\):
   \[
   4a^4 \geq 12a^2(1-2a)
   \]
   \[
   4a^4 \geq 12a^2 - 24a^3
   \]
   \[
   4a^4 + 24a^3 - 12a^2 \geq 0
   \]
   Factoring out \(4a^2\):
   \[
   4a^2(a^2 + 6a - 3) \geq 0
   \]

5. Solving the quadratic inequality \(a^2 + 6a - 3 \geq 0\):
   \[
   a = \frac{-6 \pm \sqrt{36 + 12}}{2} = \frac{-6 \pm \sqrt{48}}{2} = \frac{-6 \pm 4\sqrt{3}}{2} = -3 \pm 2\sqrt{3}
   \]
   Since \(0 < a < \frac{1}{2}\), we have:
   \[
   -3 + 2\sqrt{3} \leq a < \frac{1}{2}
   \]

6. From \(b = \frac{a}{1-2a}\) and \(b \geq a + 6\), we find the minimum values:
   \[
   r+s-rs = -3 + 2\sqrt{3}, \quad r+s+rs = 3 + 2\sqrt{3}
   \]

The final answer is \(\boxed{r+s-rs = -3 + 2\sqrt{3}, \quad r+s+rs = 3 + 2\sqrt{3}}\)