<s>[INST] Solve $ 1 - \sqrt{1 - 8\cdot(\log_{1/4}{x})^2} < 3\cdot \log_{1/4}x $ My answer: $2^\frac{-1}{\sqrt{2}} < x < 1$
Textbook answer: $2^\frac{-12}{17} < x < 1$
The only difference between my resolution and the Textbook one is that I solved by saying that
$$\text{(I) }\sqrt{1 - 8\cdot (\log_{1/4}{x})^2} >  1 -3\cdot \log_{1/4}x $$
Is true when II OR III are true:
$$\text{(II) } 1 - 3\cdot \log_{1/4}x > 0 \quad \land \quad 1-8\cdot (\log_{1/4}{x})^2 > (1 -  3\cdot \log_{1/4}x)^2  $$
$$\text{(III) } 1 - 3\cdot\log_{1/4}x < 0 \quad \land \quad 1- 8\cdot(\log_{1/4}{x})^2 > 0 $$
And the Textbook said that (I) is true when:
$$ \text{(IV) }1- 8\cdot(\log_{1/4}{x})^2 > (1 -  3\cdot \log_{1/4}x)^2 \ge 0 $$
Sorry if this is a dumb question, any help is appreciated. I can't really understand if my conditions (II or III) and their condition (IV) are both right or if one of them is wrong.
 [/INST] at first we have $$x>0$$ and $$\frac{1}{8}\geq \frac{(\ln(x))^2}{(2\ln(2))^2}$$
this condition gives us
$$e^{\ln(2)/\sqrt{2}}\geq x\geq1$$
or
$$e^{-\ln(2)/\sqrt{2}}\le x<1$$
now we can solve the inequality 
$$1+\frac{3\ln(x)}{2\ln(2)}<\sqrt{\left(1-\frac{2\ln(x)^2}{\ln(2)^2}\right)}$$
and we do case work:
I)if $$x\le e^{-\frac{2\ln(2)}{3}}$$ is hold then our inequality is true.
II) if $$x> e^{-\frac{2\ln(2)}{3}}$$ is hold then we get after squaring and simplifying
$$\frac{17}{4}(\ln(x))^2+3\ln(2)\ln(x)<0$$
from the last inequality we obtain
$$e^{-12/17\ln(2)}<x<1$$
with the condition above we get finaly $$e^{-2\ln(2)/2}<x<1$$
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