<s>[INST] How to solve $\frac{1}{2^x+3} \geq \frac{1}{2^{x+2}-1}$? I guess it's easy, but I still need help. The inequality is
$$\frac{1}{2^x+3} \geq \frac{1}{2^{x+2}-1}$$
If you set $t=2^x$, then it becomes
$$\frac{1}{t+3} \geq \frac{1}{4t-1}$$

The set of solutions is
$$x \in (-\infty,-2) \cup \{1\}$$
which is not what I get.

My attempt
$$\frac{1}{2^x+3} \geq \frac{1}{2^{x+2}-1}$$
This is undefined for $x=-2$, because the right hand side becomes: $\frac{1}{2^{-2+2}-1}=\frac{1}{2^0-1}=\frac{1}{1-1}=\frac{1}{0}$. Therefore,
$$x \neq -2$$
Now, let's set $t=2^x$.
$$\frac{1}{t+3} \geq \frac{1}{4t-1}$$
Multiply both sides with $(t+3)(4t-1)$.
We must split the inequality because we don't know is $(t+3)(4t-1)$ positive or negative. Question: what if it's zero? Should we consider that possibility too?
For $(t+3)(4t-1) > 0$:
$$4t-1 \geq t+3$$
$$3t \geq 4$$
$$t \geq \frac{4}{3}$$
$$2^x \geq \frac{4}{3}$$
$$x \geq \log_2 \left(\frac{4}{3}\right)$$
For $(t+3)(4t-1) < 0$:
$$4t-1 \leq t+3$$
$$3t \leq 4$$
$$t \leq \frac{4}{3}$$
$$2^x \leq \frac{4}{3}$$
$$x \leq \log_2 \left(\frac{4}{3}\right)$$
I think it's already obvious where and how I'm wrong, so I think I don't need to continue with this attempt. If I do, then please ask in the comment.
 [/INST] Continue from your efforts, $2^x = t$ already implies $t \gt 0$.
 $$\frac{1}{t+3}- \frac{1}{4t-1} \ge 0\\
 \frac{3t-4}{(t+3)(4t-1)}\ge 0$$
Solving this inequality with the help of zero points of the three factors, gives $t\in (-3,1/4) \cup [4/3, \infty) $ but we also had $t\gt 0$. So final answer is $t \in (0,1/4) \cup [4/3, \infty)$. 
So $x \in (-\infty, -2) \cup [2-\log_2(3), \infty)$
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