<s>[INST] What am I doing wrong solving $\sqrt{x^2+1}- 2x+1>0$? First, I started looking at where $\sqrt{x^2+1}$ is defined: $\sqrt{x^2+1}>0$ is defined everywhere. Next I 
$\sqrt{x^2+1}- 2x+1>0$
$\sqrt{x^2+1}> 2x-1$
$0> 3x^2-4x$
$0> x(3x-4)$ and I solve this for $x\in (0,\frac{4}{3}]$.
I know that this solution is wrong, because I went and drew this graph. The correct result is $x\in (-\infty,\frac{4}{3}]$. 
I'm totally confused on how to solve irrational inequalities now, because the official solving in my textbook looks like this: 
$\sqrt{x^2+1}> 2x-1$. This inequality is fulfilled , if the right side is negative, therefore $x<\frac{1}{2}$. If $x\geq\frac{1}{2}$, the right side is positive or equals to $0$ and we get $0> 3x^2-4x$ which is true for $x\in (0,\frac{4}{3}]$. Now with previous condition $x\geq\frac{1}{2}$ we get the solution $[\frac{1}{2},\frac{4}{3})$.The complete solution set is $x\in (-\infty,\frac{4}{3}]$.
I never solved this type of inequality in this way, because it's messy- Why would I look at the conditions $x<\frac{1}{2}$ and $x\geq\frac{1}{2}$ when I can immediately tell where irrational part is defined and where not?
In the end, this textbook solution process only confused me. Could anyone please explain it why the correct solution is $(-\infty,\frac{4}{3}]$ or more concretely: Where did I miss the part of the solution $(-\infty,0]$?
 [/INST] First of all, we should check for $x \ge \frac{1}{2}$ and $x < \frac{1}{2}$ because in this step:
$$\sqrt{x^2+1}> 2x-1$$
we are squaring both sides. So sign of $2x-1$ may change the direction of inequality (precisely when $|2x-1| > \sqrt{x^2+1}$).
Now, for the case where $x < \frac{1}{2}$, we also have a solution for
$$x^2+1 <4x^2-4x+1 \implies 3x^2-4x > 0 \implies x < 0 \lor x > \frac{4}{3}$$
but $x > \frac{4}{3}$ doesn't satisfy our first assumption $x < \frac{1}{2}$. Therefore, we get a solution $x < 0$ here.
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