hello, for $r+s-rs$ i have found the minimum $-2\sqrt{3}-3$ for $(r,s)=(-\sqrt{3},-\sqrt{3})$ and for $r+s+rs$ is the searched minimum $0$ for $(r,s)=(0,0)$. Sonnhard.

Sonnhard, take a moment to read the problem befoemre going to your little calculator or online program. It says positive real numbers. It's because of people like you why askers don't get the answers they want or the solutions to learn understand the solution for themselves.

hello, sorry, we have $(r+s)^2-(rs)^2=rs$ $r^2+s^2=r^2s^2-rs$ by the AM-GM inequality we get $r^2+s^2\geq rs$ $r^2s^2-rs\geq 2rs$ $r^2s^2-3rs\geq 0$ thus we obtain $rs\geq 3$ and $(r+s)^2\geq rs+(rs)^2\geq 3+9=12$ thus $r+s\geq \sqrt{12}$ and from here we get $r+s+rs\geq 2\sqrt{3}+3$ Sonnhard.

hello, for $r+s-rs=\sqrt{rs+(rs)^2}-rs>0$ this is true since $rs>0$. Sonnhard.

pythagorazz wrote: 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 $r+s+rs$ Let $r+s-rs=a,\ r+s+rs=b\Longleftrightarrow r+s=\frac{a+b}{2},\ rs=\frac{b-a}{2}.$, thus $r>0,\ s>0\Longleftrightarrow r+s>0,\ rs>0, D=(r+s)^2-4rs\geq 0$ $\Longleftrightarrow a+b>0,\ b-a>0,\ (a+b)^2\geq 8(b-a)\ \cdots [1]$. The given condition : $ab=\frac{b-a}{2}>0\ \cdots [2]$ From $[1],\ [2]$, we have $(a+b)^2\geq 8\cdot 2ab\Longleftrightarrow (a-b)^2+4ab\geq 16ab$ $\Longleftrightarrow (a-b)^2\geq 12ab$, by $[2]$, yielding $(b-a)^2\geq 6(b-a)$. Since $b-a>0$, we get $b-a\geq 6\ \cdots [3]$. Now, from $[2]$, we have $b=\frac{a}{1-2a}>0$, thus substituting this $[3]$ gives $\frac{a}{1-2a} -a\geq 6$ with $0<a<\frac 12$, or $a^2+6a-3\geq 0$ and $0<a<\frac 12\ \cdots [4]$ $-3+2\sqrt{3}<\frac 12\Longleftrightarrow 2\sqrt{3}<\frac 72\Longleftrightarrow \sqrt{48}<\sqrt{49}$, so from $[4]$, yielding $-3+2\sqrt{3}\leq a<\frac 12.$ Therefore, by $[3]\Longleftrightarrow b\geq a+6$, the desired minimum values are : $r+s-rs=-3+2\sqrt{3},\ r+s+rs=3+2\sqrt{3}.$

hello, kunny have right with $s=1/2\,{\frac { \left( 1+\sqrt {-3+4\,{r}^{2}} \right) r}{-1+{r}^{2}}}$ we get $-1/2\,{\frac { \left( -2\,r-1+\sqrt {-3+4\,{r}^{2}} \right) r}{r+1}}\geq 2\sqrt{3}-3$ Sonnhard.

$rs=(r+s)^2-(rs)^2\ge 4rs-(rs)^2 \Rightarrow rs\ge 3$ $r+s+rs=\sqrt{(rs)^2+rs}+rs\ge\sqrt{3^2+3}+3=2\sqrt{3}+3$. $r+s-rs=\sqrt{(rs)^2+rs}-rs=\frac{1}{\sqrt{1+\frac{1}{rs}}+1}\ge \frac{1}{\sqrt{1+\frac{1}{3}}+1}=2\sqrt{3}-3$.

Wow, that's very simple solution, sqing.

kunny wrote: Wow, that's very simple solution, sqing. Happy lunar New Year .