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In the paper linear forms in the logarithms of real algebraic numbers close to 1, it is written on page 8 that- $$ ab^2 \geq k(ab)^\frac{2(k-1)}{k} \cdots (1)$$ How we prove above above inequality? We know that- $(a+1)(ab^2+1) \geq ((ab+1)^\frac{2}{k}+1)^k \implies (a+1)(ab^2+1) \geq (ab+1)^2+ k(ab+1)^{2(k-1)/k} \implies ab^2+a \geq 2ab+ k(ab+1)^{2(k-1)/k} $, but how do we get $ ab^2 \geq k(ab)^\frac{2(k-1)}{k}?$

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We know that- $(a+1)(ab^2+1) \geq ((ab+1)^\frac{2}{k}+1)^k$ $\implies (a+1)(ab^2+1) \geq (ab+1)^2+ k(ab+1)^{2(k-1)/k}$

[Note,we are using only two terms of binomial expansion of $((ab+1)^\frac{2}{k}+1)^k$]

$\implies ab^2+a \geq 2ab+ k(ab+1)^{2(k-1)/k} $

$\implies ab^2 \geq 2ab-a + k(\sum_{m=0}^{\frac{2(k-1)}{k}}\binom{2(k-1)/k}{ m} (ab)^{m}) $

$\implies ab^2 \geq 2ab-a + k(\binom{2(k-1)/k)}{ 2(k-1)/k)} (ab)^\frac{2(k-1)}{k} + \sum_{m=0}^{\frac{2(k-1)}{k}-1}\binom{(2(k-1)/k)-1}{ m} (ab)^{m}) $

$\implies ab^2 \geq 2ab-a + k (ab)^\frac{2(k-1)}{k} + k( \sum_{m=0}^{\frac{2(k-1)}{k}-1}\binom{(2(k-1)/k)-1}{ m} (ab)^{m}) $

$\implies ab^2 \geq k (ab)^\frac{2(k-1)}{k} $ [Note, $2ab-a>0$]

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