<s>[INST] Help: $ |\frac{a+1}{a}- (\frac{xz}{y^2})^k|\leq \frac{1}{b}$ 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?
$$ \left|\frac{a+1}{a}- \frac{(a+1)(ab^2+1)}{(ab+1)^2}\right|\leq \frac{1}{b}$$ 
 [/INST] 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 $$b\geq 2,a\geq 2^{29}$$
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