I will begin with my original problem: Let $r : [0,1) \to (0,\infty )$ a continuous, increasing function such that $\lim_{h\to 1} r(h) = \infty$. In my case $r$ is explicitly given. For fixed $x\in (0,1)$ the set of $h$ fulfilling $$x^{r(h) - r(0)} + x^{8r(h)} - 1 \geq 0$$ is of the form $[0, h_0 (x)]$. One can see that $\lim_{x\to 0} h_0 (x) = 0$. Actually, I want to say something about the speed of this convergence. For this purpose I tried to consider the problem the other way round.
This is, for fixed $h\in [0,1)$ one can try to bound the unique root $x_0 (h) := x_0 (a,b) \in (0,1)$ of
$$p_{a,b} (x) := x^b + x^a -1$$
in terms of $b := r(h) - r(0)$, $a := 8r(h)$. Indeed, assume $B(r(h))$ is a lower bound. That is $B(r(h)) \leq x_0 (h)$. Then $B( r(h_0 (x)) ) \leq x_0 (h_0 (x)) = x$. If $B$ is nice enough and increasing, we will get information about the speed of convergence.
My question is how to bound $x_0 (a,b)$ from below.
I tried the following: For $a \leq a' , b \leq b' $ we have $$p_{a,b} \geq p_{a',b'}$$ which implies that $x_0 (a,b) \leq x_0 (a',b')$. Thus I had the idea to define $\lfloor s \rfloor_n := \sup\{ \frac k n \leq s : k\in \Bbb{N}_0 \}$. Then $x_0 (\lfloor a\rfloor_n , \lfloor b\rfloor_n) \leq x_0 (a,b)$. The advantage is that $$p_{\lfloor a\rfloor_n , \lfloor b\rfloor_n} (z^n) = z^{n \cdot \lfloor b\rfloor_n } + z^{n \cdot \lfloor a\rfloor_n } -1$$ is an ordinary polynomial in $z$. In particular it is an trinomial with coefficients $1,1,-1$. The naive way to bound the roots of $z^m + z^l -1$ from below is to search for a upperbound of the roots of $-z^{\max (m,l)} + z^{\vert m -l\vert} + 1$ (cf. https://en.wikipedia.org/wiki/Geometrical_properties_of_polynomial_roots#Bounds_on_all_roots) and take the inverse. By Chauchy's bound and Lagrange's bound this would result in a lower bound $x_0 (\lfloor a\rfloor_n , \lfloor b\rfloor_n) \geq (\frac 1 2 ) ^n \to 0$, which does not help much. This is, because the lower bound does not involve the degrees. One also sees that this is a very bad lower bound because the root of $z^m + z^l -1$ in $(0,1)$ goes to $1$ if $m,l$ are growing large.
Is there a possibility to bound the root of $z^m + z^l -1$ in $(0,1)$ from below in terms of the degrees? Any reference on bounding positive roots is also appreciated.
Edit: For $h$ being small $b$ will be small eventually. Martin R's comment suggested the bound $x_0(a,b) \geq 2^{-\frac 1 b}$. This can be achieved by the following: For $a-b \geq 0$ root $x$ will fulfil $x^b (1 + x^{a-b}) = 1$. Since $x^{a-b} \in (0,1)$ we have $2x^b \geq x^b (1 + x^{a-b}) = 1$, thus $x^b \geq 2^{-\frac 1 b}$.
For the idea formulated above this leads to $r(h_0 (x) ) - r(0) \leq -\frac{\log (2)}{\log (x)}$.
I do not know if this will give the good bounds you look for.
Admitting that $(m,l)$ are large, the solutions are close to $1$. So, let $z=1-x$ and expand as series around $x=0$. This would give $$(1-x)^m+(1-x)^l-1=1- (l+m)x+\frac{1}{2} ((l-1) l+(m-1) m)x^2+-\frac{1}{6} \left( (l-2) (l-1) l- (m-2) (m-1) m\right)x^3+O\left(x^4\right)$$ Now, using series reversion, this would give, as an approximation, $$x=\frac{1}{l+m}+\frac{(l-1) l+(m-1) m}{2 (l+m)^3}+O\left(\frac 1{(l+m)^5} \right)$$ Trying for $m=12$ and $l=34$, this would give $x=\frac{2743}{97336}$ that is to say $z=\frac{94593}{97336}\approx 0.9718$ while the "exact" solution would be $0.9676$.
For the reversed series, I did not write the next term which is quite messy but it seems to me that, if $l > m$ it is positive. So, the $x$ will be a lower bound and the corresponding $z$ an upper bound.
Using the next (not written term), for the worked example, we would have $x=\frac{3152531}{102981488}$ that is to say $z=\frac{99828957}{102981488}\approx 0.9694$ which is much better.