All right. The hardest thing is to do all algebra right (I am always prone to misplacing + and -, so check everything I say in the first part. :)).
Let $\lambda=\log \frac 1p$ and $\Gamma=\log\frac 1s$. We want to create a situation when there are 3 points on the line $D=1+br$ ($b>0$). Let's write $(ps^r)^{D/D+r}$ as the exponent of minus $\lambda\frac{1+br}{1+(b+1)r}+\Gamma\frac{r(1+br)}{1+(b+1)r}$. Our first step will be to replace $(b+1)r$ by $r$. Since $\Gamma>0$ is free ($\lambda$'s are restricted to $\sum_j e^{-\lambda_j}=1$), we still get $\lambda\frac{1+br}{1+r}+\Gamma\frac{r(1+br)}{1+r}$ but now $b\in(0,1)$. Now replace $1+r$ with $r$. We'll get $\lambda\frac{1-b+br}{r}+\Gamma\frac{(r-1)(1-b+br)}{r}$. Now let's open the parentheses and group the terms. If I haven't made an odd number of errors, we get $(\lambda-\Gamma)(1-b)r^{-1}+\Gamma b r+(\lambda-\Gamma)b+\Gamma(1-b)$ ($r\ge 1$). Now let us replace $r$ by $\frac{1-b}{b}r$ to get $(\lambda-\Gamma)br^{-1}+\Gamma(1-b) r+(\lambda-\Gamma)b+\Gamma(1-b)$ ($r\ge \frac b{1-b}$)and denote $X=(\lambda-\Gamma)b$, $Y=\Gamma(1-b)$. Thus, the problem is reduced to asking whether the sum of two exponents of the kind $$ \exp(-Xr^{-1}-Yr-X-Y) $$ where $r> 0$, $Y>0$ and $X$ is unrestricted can take the value $1$ three times on the positive semiaxis (the leftmost root will be $b/(1-b)$ after which you can go back and discern the initial values).
The rest is simple analysis.
Let for one exponent $X=Y=B$. Then at $1$, it is $e^{-4B}$ and at $1/2$ it is $e^{-4.5B}$. Now for the second exponent choose $Y=A$ and $X=-\varepsilon$. Then $X$ guarantees that we have $+\infty$ at $0+$ but is invisible for any noticeable positive $r$. Also, at $+\infty$, we have $0$, so we just want the value at $1$ to be larger than $1$ and the value at $1/2$ to be smaller than $1$, which results in the system of inequalities $$ e^{-1.5A}+e^{-4.5B}<1;\qquad e^{-2A}+e^{-4B}>1 $$ Now, take relatively small $A$ and put $e^{-4B}=2A$. The second inequality is then fine and the first one is $e^{-1.5 A}+A^{9/8}<1$, which is true if $A$ is small enough.