How to prove $b \ll a (a^{2}b^2)^{2.1/k} $ 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.

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}} $
