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I have to show that some set of ODE's has a unique fix point which boils down to proving that for any $\lambda > 0$, $B \in \mathbb{N}_0$ and $r >0$ we have that the maps: $$ f_1(x) :=x - 1 + (1-\lambda)\frac{1 - \left( \frac{\lambda}{1-x} + rx \right)^{B+1}}{1 - \left( \frac{\lambda}{1-x} + rx \right)} $$ and $$ f_2(x) := \left( \frac{\lambda + rx(1-x)}{1-x} \right) \cdot \left( (1-\lambda) \left(\lambda + \frac{\lambda x}{1-x} + rx\right)^B \right) + rx^2 - rx -x $$ share one unique root (i.e. $\exists! x \in [0,\infty[: f_1(x) = f_2(x) = 0$).

I have done numeric experiments which seem to show that $f_1$ has a unique root on $[0,\infty)$, moreover this root seems to always be a root of $f_2$ as well (this is of course not a proof, so counterexamples are also welcome).

I tried to show that $f_1$ is convex, it obviously suffices to show that $g(x) := \frac{1 - \left( \frac{\lambda}{1-x} + rx \right)^{B+1}}{1 - \left( \frac{\lambda}{1-x} + rx \right)}$ is convex, i.e. $g''(x) \geq 0$, but this is unfortunately not true in general so another path has to be taken.

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    Have you tried to check that $f_1(0) < 0$, $f_1 (+\infty) > 0$ and $f_1'(x) > 0$ when $x \in \lbrack 0, +\infty)$ ? – 2017-02-06
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    What a coincidence! I was doing that now, but at the moment I'm stuck on showing that $(n-1)x^n \geq nx^{n-1} - 1$ – 2017-02-06
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    Okay, got it with induction. I think this solved my problem (just need to properly write everything down) – 2017-02-06
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    Technically you can write this as an answer to your post :) – 2017-02-06

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One can show that $f_1$ is ascending on $[0,1[$, strictly negative in $0$ and infinity at $1$, it is clear that $f_1$ is strictly positive, which shows that $f_1$ has a unique root. Moreover one can find a polynomial $h$ s.t. $f_2 = h \cdot f_1$ showing that the unique root of $f_1$ is also a root of $f_2$.