Proof-Explanation : $(a+1)(ab^2+1) \leq b^3$ In the paper linear forms in the logarithms of real algebraic numbers close to 1, it is written on page $9$ that-
$(xz)^k\leq b^3 \implies (a+1)(ab^2+1) \leq b^3$ , I could not figure it out.
Here, $(xz)^k=(a+1)(ab^2+1)$ (see page $8$),$a, b \geq 2, $ and both are integers, $k\geq 3 $ (see theorem 11 on page $8$, please see  the attached-paper for detail ).
If we simplify by Rearrangement inequality, we have that
$$(a+1)(ab^2+1)=a^2b^2+ab^2+a+1 $$
how this could be less than or equal to $b^3$?

On page 8 there is the inequality
$$ b\geq (k^{k}a^{k-2})^{1/2} \label{12}\tag{12} $$
and due to the comment on page 9 we may assume that $k\geq 50$ and $a\geq 2^{49}$. Since $b\geq 1$, we see by \eqref{12} that
$$b^{3}\geq b^{2}(k^{k}a^{k-2})^{1/2} \label{*}\tag{*}$$
Now we have, since $b\geq 1$, that
\begin{align}
(xz)^{k}&=&(a+1)(ab^2+1)\\
&=&a^2b^2+a+ab^2+1\\
&\leq&b^2(a^2+a+a+1)\\
&=&b^2(a+1)^2.
\end{align}
Now we note that
$$(a+1)^2=a^2+2a+1\leq a^2+2a^2+a^2=4a^2,$$
whence we see that
$$(a+1)^{2}\leq 4a^{2}\leq k^{k/2}a^{\frac{k-2}{2}}$$
since $k\geq 50$. Therefore we have by \eqref{*} that
$$(xy)^{k}\leq b^{2}(a+1)^{2}\leq b^{2}(k^{k}a^{k-2})^{1/2}\leq b^{3},$$
as desired
