# Help: $|\frac{a+1}{a}- (\frac{xz}{y^2})^k|\leq \frac{1}{b}$

In the paper linear forms in the logarithms of real algebraic numbers close to 1, it is written on page $9$ that

On other hand, a short calculation yields $$\left|\frac{a+1}{a}- \left(\frac{xz}{y^2}\right)^k\right|\leq \frac{1}{b}$$

Image of the page :- Here, $$\left(\frac{xz}{y^2}\right)^k= \frac{(a+1)(ab^2+1)}{(ab+1)^2}$$ and $b \geq 2, a\geq 2^{49},k\geq 50$ (see page $8, 9$).

So, how do we prove the following?

$$\left|\frac{a+1}{a}- \frac{(a+1)(ab^2+1)}{(ab+1)^2}\right|\leq \frac{1}{b}$$

We have, using $b\ge 2$ and $a\ge 2^{49}$, \begin{align}\left|\frac{a+1}{a}- \frac{(a+1)(ab^2+1)}{(ab+1)^2}\right| &=(a+1)\left|\frac{1}{a}- \frac{(ab^2+1)}{(ab+1)^2}\right| \\\\&=(a+1)\left|\frac{(ab+1)^2-a(ab^2+1)}{a(ab+1)^2}\right| \\\\&=(a+1)\left|\frac{1+a(2b-1)}{a(ab+1)^2}\right| \\\\&=(a+1)\cdot \frac{1+a(2b-1)}{a(ab+1)^2} \end{align}

Here, since $$ab+1\ge ab$$ we have $$\frac{1}{ab+1}\color{red}{\le}\frac{1}{ab}$$ Also, we have $$a+1\le \frac{a^2}{2}\quad\text{and}\quad 1+a(2b-1)\le 2ab$$

Using these gives $$\left|\frac{a+1}{a}- \frac{(a+1)(ab^2+1)}{(ab+1)^2}\right| =(a+1)\cdot \frac{1+a(2b-1)}{a(ab+1)^2}\le \frac{a^2}{2}\cdot\frac{2ab}{a(ab)^2}=\frac 1b$$

• When u divide something with big nunber the amount become smaller, I urge u to check/revise last paragraph of your answer – Michael Sep 11 '18 at 3:48
• @Mike SQ : The last inequality is true since $ab+1\ge ab$ implies $\frac{1}{ab+1}\color{red}{\le}\frac{1}{ab}$. I've added some explanations in my answer. – mathlove Sep 11 '18 at 5:50
• @Mike SQ : I'm going to give another explanation. After having $(a+1)\cdot \frac{1+a(2b-1)}{a}\le\frac{a^2}{2}\cdot \frac{2ab}{a}$, which should be clear, we divide the LHS with larger number $(ab+1)^2$ and divide the RHS with smaller number $(ab)^2$, so we get our last inequality. I hope this helps. – mathlove Sep 11 '18 at 6:36

I have got $$\left|\frac{a+1}{a}-\frac{(a+1)(ab^2+1)}{(ab+1)^2}\right|=\left|{\frac { \left( a+1 \right) \left( 2\,ab-a+1 \right) }{a \left( ab+1 \right) ^{2}}} \right|$$ I have compute $$\frac{1}{4}-f(a,b)^2=\frac{\left(a^3 b^2-2 a^2 b+2 a^2-4 a b+a-2\right) \left(a^3 b^2+6 a^2 b-2 a^2+4 a b+a+2\right)}{4 a^2 (a b+1)^4}$$ and this is positive if $$b\geq 2,a\geq 2^{29}$$

• Your inequality is right, defining $$f(a,b)=\left|{\frac { \left( a+1 \right) \left( 2\,ab-a+1 \right) }{a \left( ab+1 \right) ^{2}}} \right|$$ and you must searched for a maximum of $$f(a,b)$$ – Dr. Sonnhard Graubner Sep 8 '18 at 10:37
• ... need a proof !!! :) I could not find one.... – Michael Sep 10 '18 at 9:51