Vtyjkyuk

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
$$ left|fraca+1a- left(fracxzy^2right)^kright|leq frac1b$$

Image of the page :-

Here,
$$left(fracxzy^2right)^k= frac(a+1)(ab^2+1)(ab+1)^2$$ and $ b geq 2, ageq 2^49,kgeq 50 $ (see page $8, 9$).

So, how do we prove the following?

$$ left|fraca+1a- frac(a+1)(ab^2+1)(ab+1)^2right|leq frac1b$$

We have, using $bge 2$ and $age 2^49$,
$$beginalignleft|fraca+1a- frac(a+1)(ab^2+1)(ab+1)^2right|
&=(a+1)left|frac1a- frac(ab^2+1)(ab+1)^2right|
\\&=(a+1)left|frac(ab+1)^2-a(ab^2+1)a(ab+1)^2right|
\\&=(a+1)left|frac1+a(2b-1)a(ab+1)^2right|
\\&=(a+1)cdot frac1+a(2b-1)a(ab+1)^2
endalign$$

Here, since
$$ab+1ge ab$$
we have
$$frac1ab+1colorredlefrac1ab$$
Also, we have
$$a+1le fraca^22quadtextandquad 1+a(2b-1)le 2ab$$

Using these gives
$$left|fraca+1a- frac(a+1)(ab^2+1)(ab+1)^2right|
=(a+1)cdot frac1+a(2b-1)a(ab+1)^2le fraca^22cdotfrac2aba(ab)^2=frac 1b$$

I have got $$left|fraca+1a-frac(a+1)(ab^2+1)(ab+1)^2right|=left|frac left( a+1 right) left( 2,ab-a+1 right) a left( ab+1
right) ^2
right|$$
I have compute $$frac14-f(a,b)^2=fracleft(a^3 b^2-2 a^2 b+2 a^2-4 a
b+a-2right) left(a^3 b^2+6 a^2 b-2 a^2+4 a
b+a+2right)4 a^2 (a b+1)^4$$ and this is positive if $$bgeq 2,ageq 2^29$$