Help: $ |\frac{a+1}{a}- (\frac{xz}{y^2})^k|\leq \frac{1}{b}$

Question

In the paper linear forms in the logarithms of real algebraic numbers close to 1, it is written on page $9$ that

On other hand, a short calculation yields $$ \left|\frac{a+1}{a}- \left(\frac{xz}{y^2}\right)^k\right|\leq \frac{1}{b}$$

Image of the page :-

Here, $$\left(\frac{xz}{y^2}\right)^k= \frac{(a+1)(ab^2+1)}{(ab+1)^2}$$ and $ b \geq 2, a\geq 2^{49},k\geq 50 $ (see page $8, 9$).

So, how do we prove the following?

$$ \left|\frac{a+1}{a}- \frac{(a+1)(ab^2+1)}{(ab+1)^2}\right|\leq \frac{1}{b}$$

Answer 1

We have, using $b\ge 2$ and $a\ge 2^{49}$, $$\begin{align}\left|\frac{a+1}{a}- \frac{(a+1)(ab^2+1)}{(ab+1)^2}\right| &=(a+1)\left|\frac{1}{a}- \frac{(ab^2+1)}{(ab+1)^2}\right| \\\\&=(a+1)\left|\frac{(ab+1)^2-a(ab^2+1)}{a(ab+1)^2}\right| \\\\&=(a+1)\left|\frac{1+a(2b-1)}{a(ab+1)^2}\right| \\\\&=(a+1)\cdot \frac{1+a(2b-1)}{a(ab+1)^2} \end{align}$$

Here, since $$ab+1\ge ab$$ we have $$\frac{1}{ab+1}\color{red}{\le}\frac{1}{ab}$$ Also, we have $$a+1\le \frac{a^2}{2}\quad\text{and}\quad 1+a(2b-1)\le 2ab$$

Using these gives $$\left|\frac{a+1}{a}- \frac{(a+1)(ab^2+1)}{(ab+1)^2}\right| =(a+1)\cdot \frac{1+a(2b-1)}{a(ab+1)^2}\le \frac{a^2}{2}\cdot\frac{2ab}{a(ab)^2}=\frac 1b$$

Answer 2

I have got $$\left|\frac{a+1}{a}-\frac{(a+1)(ab^2+1)}{(ab+1)^2}\right|=\left|{\frac { \left( a+1 \right) \left( 2\,ab-a+1 \right) }{a \left( ab+1 \right) ^{2}}} \right|$$ I have compute $$\frac{1}{4}-f(a,b)^2=\frac{\left(a^3 b^2-2 a^2 b+2 a^2-4 a b+a-2\right) \left(a^3 b^2+6 a^2 b-2 a^2+4 a b+a+2\right)}{4 a^2 (a b+1)^4}$$ and this is positive if $$b\geq 2,a\geq 2^{29}$$