In the paper linear forms in the logarithms of real algebraic numbers close to 1, it is
written on page $9$ that
It follows from (13) and Lemma 2 with $\eta=0.1 $ we have that, $$b \ll a (a^{2}b^2)^{2.1/k} $$
The image of page 9 :
How do we prove $b \ll a (a^{2}b^2)^{2.1/k} $?
My Confusion : Note, if $k$ increases then ${2.1/k}$ goes close to $0$, so it is confusing
how the inequality holds true.
On
We know, $ \left|\frac{a+1}{a}- \left(\frac{xz}{y^2}\right)^k\right|\leq \frac{1}{b} \cdots (13)$
Here, $xz= ((a+1)(ab^2+1))^\frac{1}{k}, y^2= (ab+1)^\frac{2}{k}.$
So, $ \left|\frac{a+1}{a}- \frac{(xz)^k}{(ab+1)^2} \right|\leq \frac{1}{b} \cdots (a)$
Since, $(ab+1)>ab\implies(ab+1)^\frac{2}{k}>(ab)^\frac{2}{k}$
$\implies \frac{1}{(ab+1)^\frac{2}{k}}<\frac{1}{(ab)^\frac{2}{k}} \implies \frac{(xz)^k}{(ab+1)^\frac{2}{k}}<\frac{(xz)^k}{(ab)^\frac{2}{k}}$
$\implies -\frac{(xz)^k}{(ab+1)^\frac{2}{k}}> -\frac{(xz)^k}{(ab)^\frac{2}{k}}$ [since, the inequality sign
changes when both sides are multiplied by $-1$.]
$\implies-\frac{(xz)^k}{(ab)^\frac{2}{k}} < -\frac{(xz)^k}{(ab+1)^\frac{2}{k}}$ [changing sides]
$\implies (\frac{a+1}{a}) -\frac{(xz)^k}{(ab)^\frac{2}{k}} < (\frac{a+1}{a}) -\frac{(xz)^k}{(ab+1)^\frac{2}{k}}$
$\implies \left|\frac{a+1}{a}- \frac{(xz)^k}{(ab)^\frac{2}{k}} \right| \leq \left|\frac{a+1}{a}- \frac{(xz)^k}{(ab+1)^\frac{2}{k}} \right|$
Using inequality (a), we deduce-
$\left|\frac{a+1}{a}- \frac{(xz)^k}{(ab)^\frac{2}{k}} \right| \leq \frac{1}{b} \cdots (b)$
By inspection, we see- $ \left|(\frac{a+1}{a})^\frac{1}{k} \right |\leq \left|\frac{a+1}{a} \right | $
$\implies \left|(\frac{a+1}{a})^\frac{1}{k}- \frac{(xz)^k}{(ab)^\frac{2}{k}}\right |\leq \left|\frac{a+1}{a}- \frac{(xz)^k}{(ab)^\frac{2}{k}} \right | $
[subtracting $\frac{(xz)^k}{(ab)^\frac{2}{k}}$ on both sides]
Using inequality (b), we deduce-
$\left|(\frac{a+1}{a})^\frac{1}{k}- \frac{(xz)^k}{(ab)^\frac{2}{k}}\right |\leq \frac{1}{b}\ \cdots (c) $
To use Lemma 2, we need to re-parameterize $p, q$ of Lemma 2
according to inequality $(13)$, so let, $p= (xz)^k, q= (ab)^\frac{2}{k} $, then-
$ \left|(1+\frac{1}{a})^\frac{1}{k}- \left(\frac{p}{q}\right)\right| > \frac{c_5}{aq^{\epsilon +2 }} $ [ Lemma 2]
$\implies \left|(1+ \frac{1}{a})^\frac{1}{k}- \frac{(xz)^k}{(ab)^\frac{2}{k}} \right| > \frac{c_5}{a(ab)^{\frac{2(\epsilon +2 )}{k}}} $
$\implies \frac{1}{b} > \frac{c_5}{a(ab)^{\frac{2(\epsilon +2 )}{k}}} $ [ using inequality (c)]
$\implies a(a^2b^2)^{\frac{2.1}{k}} >bc_5 $ [$\epsilon =0.1$]
$\implies a(a^2b^2)^{\frac{2.1}{k}} >b \implies b \ll_k a(a^2b^2)^{\frac{2.1}{k}} $
This is a partial solution to what I think is your problem (namely, that you're having a tough time working through this paper), although not to the immediate question you asked; I'm suggesting that you consider writing to the author saying something like this:
It's hard for me to imagine that you'd get no response at all to such a note, even if the response is something like "I think, judging by some of your earlier questions, that perhaps that this paper is a bit too advanced for your current level of mathematical understanding and sophistication. Perhaps you should complete at least the junior-level courses for mathematics majors at your university before trying to read it, or you'll have troubles with almost every sentence." The author, after all, is probably very well suited to judge what preparation is needed to read his paper.
What I anticipate is that the author will say something like "This statement holds under the hypotheses written above; perhaps I should have been clearer, but $a$ must be at least ..., and $k$ is no greater than ..., hence ..."