<s>[INST] Compare $A=\frac{1.0\,000\,004}{(1.0\,000\,006)^2}$ and $ B=\frac{(0.9\,999\,995)^2}{0.9\,999\,998}$ My work:


*

*$1.0\,000\,004 = 1+\frac{4}{10^7}=1+\frac{1}{125\cdot 10^6}$

*$ (1.0\,000\,006)^2=(1+\frac{6}{10^7})^2=(1+\frac{3}{5\cdot 10^6})^2$

*$ (0.9\,999\,995)^2=(\frac{9\,999\,995}{10^7})^2=(\frac{1\,999\,999}{2\cdot10^6})^2$

*$ 0.9\,999\,998=\frac{9.999\,998}{10^7}=\frac{4\,999\,999}{5\cdot10^6}$
The two fractions become:
$A=\dfrac{1+\frac{1}{125\cdot 10^6}}{(1+\frac{3}{5\cdot 10^6})^2}$ and $B=\dfrac{(\frac{1\,999\,999}{2\cdot10^6})^2}{\frac{4\,999\,999}{5\cdot10^6}}$
At this stage, I don't know how to continue.
Any help is appreciated. Thank you.
 [/INST] Let $a=\frac{1}{10^7}$.
Then, $$A=\frac{1+4a}{(1+6a)^2},\quad B=\frac{(1-5a)^2}{1-2a}$$
Now, 
$$\begin{align}A-B&=\frac{1+4a}{(1+6a)^2}-\frac{(1-5a)^2}{1-2a}\\&=\frac{(1+4a)(1-2a)-(1-5a)^2(1+6a)^2}{(1+6a)^2(1-2a)}\\&=\frac{3 a^2 (20a (1-15 a)+17)}{(1+6a)^2(1-2a)}\end{align}$$
This is positive, so $A\gt B$.
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