Show there exists a $a>0$ such that $$e^{-s(a+r)}<\sqrt{(1-rs)^2-s^2}$$ where $s=\frac{-1-r^2-2ar+\sqrt{(1-r)^2+4(a+r)^2}}{2(1-r^2)(a+r)}\geq0$ and $-1\leq r\leq1$.
This is a part of a bigger problem. I can solve it when $r=0$, in which case we have to prove: $$e^{-2sa}<1-s^2$$ where $s=\frac{-1+\sqrt{1+4a^2}}{2a}$. We can choose $a$ small enough such that $0<sa<1.5$. Since $e^{-x}<1-x/2 $ for $0<x<1.5$, $$e^{-2sa}<1-sa < 1-s^2,$$ where the last step can easily be verified. Any idea for the general case?
Hint: With the Substitution $$t=\frac{-1+\sqrt{1+4a}}{2}$$ you will get $$e^{-t}<1-\frac{1}{2}t$$. I think that is easy to solve.