Infinite inequality with logarithm

Solve inequality $3 - \log_{0.5}x - \log^2_{0.5}x - \log^3_{0.5}x - \cdots \ge 4\log_{0.5}x$ Any suggestions how to start?

• Let $u = \log_{0.5} x$. Find the admissible $u$. – Daniel Fischer Oct 10 '15 at 19:57
• Then we have $3 - u - u^2 - u^3 - ... \ge 4u$. Still, have no idea what to do next. – user128409235 Oct 10 '15 at 20:21

We have

\begin{align} 3-\log_{0.5}x-\log^2_{0.5}x-\log^3_{0.5}x-\cdots&=4-\sum_{n=0}^{\infty}\log^n_{0.5}x\\\\ &=4-\frac{1}{1-\log_{0.5}x}\\\\ &=\frac{3-4\log_{0.5}x}{1-\log_{0.5}x} \end{align}

for $|\log_{0.5}x|<1\implies 0.5<x<2$. Now note that the inequality

$$\frac{3-4\log_{0.5}x}{1-\log_{0.5}x}\ge 4\log_{0.5}x$$

is equivalent to

$$\left(2\log_{0.5}x-3\right)\left(2\log_{0.5}x-1\right)\ge 0$$

which implies $\log_{0.5}x\le 0.5$ or $x\ge \sqrt{2}/2$. Therefore, we have

$$\bbox[5px,border:2px solid #C0A000]{\sqrt{2}/2\le x<2}$$

• Thank you. I forogt about formula for sum of infinite geometric series. – user128409235 Oct 10 '15 at 20:23
• You're welcome. My pleasure. Please check to ensure that these inequalities make sense to you. – Mark Viola Oct 10 '15 at 20:24
• Yes, I understand them. – user128409235 Oct 10 '15 at 20:30
• Excellent! Pleased to hear. – Mark Viola Oct 10 '15 at 20:31