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-

$$(a+1)(ab^2+1) \geq (ab+1)^2+ \sum_{n=0}^{k-1} \binom{k-1}{n} (ab+1)^\frac{2n}{k} \cdots (2) ?$$

• @Imago then i have to show $2ab+3$ is greater than that sum??? ... not clear yet!! :) – Mike SQ Sep 6 '18 at 13:35
• My first guess fails :/, it looked to me, one could just apply the Bernoulli inequality.., but I guess it needs some more work. – Imago Sep 6 '18 at 13:41
• @Imago the way it is presented, I thought it was trivial and i was the only person who couldn't see !! :) – Mike SQ Sep 6 '18 at 13:47

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

$$(a+1)(ab^2+1) \geq ((ab+1)^\frac{2}{k}+1)^k$$