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-

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

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

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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.