<s>[INST] How to prove $(a+1)(ab^2+1) \geq ((ab+1)^\frac{2}{k}+1)^k$? 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-
$$(a+1)(ab^2+1) \geq (ab+1)^2+ \sum_{n=0}^{k-1} \binom{k-1}{n} (ab+1)^\frac{2n}{k} \cdots (2) ?$$ 
 [/INST] 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
$$(a+1)(ab^2+1) \geq ((ab+1)^\frac{2}{k}+1)^k$$
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