I came across this problem:
$$\sqrt{(x^2+1)}-2x+1>0$$
So I wanted to solve it by squaring the inequality:
$$\sqrt{(x^2+1)}>2x-1$$
If I understand correctly if I want to square the equation, I need to show that $2x-1$ is a positive number, as $\sqrt{(x^2+1)}$ is positive for all real numbers.
So I got $2x-1\geqslant 0$ , so we get: $x\geqslant\frac{1}{2}$
I used this as a perimeter for the first equation when inequalities do not change, after squaring both sides I got:
$$(x^2+1)>4x^2-4x+1$$
$$0>x(3x-4)$$
Then we can see that the equation is less than zero on interval: $(0,\frac{4}{3})$
Here I used intersection with $x\geqslant\frac{1}{2}$.
So the parameter becomes(Solution 1): $[\frac{1}{2},\frac{4}{3})$
Then I created another parameter for when $x<\frac{1}{2}$
After some algebra with the same square but the change of the direction of inequality, I got the new parameter intersection (Solution 2) with $x<\frac{1}{2}$ :
$$x<0$$
By joining the solution 1 and solution 2 with union, I got parameter for $x$:
$$(-\infty,0) \cup [\frac{1}{2}, \frac{4}{3}) $$
There seems to be something that I missed, as the correct solution is: $x<\frac{4}{3}$
Where did my problem-solving go wrong?
Best Answer
Since $x^2+1>0$, the inequality
$$\sqrt{x^2+1}>2x-1$$
is always satisfied for $2x-1\le 0 \implies x\le \frac12$.
Then for $2x-1>0$ we can square to obtain
$$x^2+1>4x^2-4x+1 \iff3x^2-4x<0 \iff 0<x<\frac43$$
and we need to take the solutions compatible with the condition $2x-1>0$ that is $x>\frac12$.
Therefore solutions are
that is $x<\frac43$.