In the paper linear forms in the logarithms of real algebraic numbers close to 1, it is
written on page 8 that-
$$(a+1)(ab^2+1) \geq ((ab+1)^\frac{2}{k}+1)^k=(ab+1)^2+ \sum_{n=0}^{k-1} \binom{k-1}{n} (ab+1)^\frac{2n}{k} \cdots (1)$$
Clearly, $(a+1)(ab^2+1) > (ab+1)^2$, but right hand side of equation (1) has a sum $\sum_{n=0}^{k-1} \binom{k-1}{n} (ab+1)^\frac{2n}{k}$, so how we prove that-
$$(a+1)(ab^2+1) \geq (ab+1)^2+ \sum_{n=0}^{k-1} \binom{k-1}{n} (ab+1)^\frac{2n}{k} \cdots (2) ?$$
The inequality is obtained in two steps.
At first, for example by Rearrangement inequality, we have that
$$(a+1)(ab^2+1)=a^2b^2+ab\cdot b+a\cdot 1+1\ge a^2b^2+ab\cdot 1+a\cdot b+1$$
$$\ge a^2b^2+2ab+1= (ab+1)^2$$
but inequality holds $\iff$ $b=1$ and since $b>1$ we have that
$$(a+1)(ab^2+1)\color{red}> (ab+1)^2$$
then, second step, since the numbers are perfect $k$ powers we have that
$$(a+1)^{1/k}(ab^2+1)^{1/k}>(ab+1)^{2/k} \implies (a+1)^{1/k}(ab^2+1)^{1/k}\ge(ab+1)^{2/k}+1$$
that is
$$(a+1)(ab^2+1) \geq ((ab+1)^\frac{2}{k}+1)^k$$