How to find a solution for this inequation? what's the best way to find a solution for the following inequation:
$$
\sqrt{x^2-1}>x
$$
The result is as Wolfram says:
$$
x \leq-1
$$
 A: The main thing to note is that you mustn't directly square the terms.


*

*In order that the inequality makes sense, you need $x^2-1\ge0$

*Recall that $\sqrt{a}\ge0$ (when $a\ge0$) by definition, so an inequality $\sqrt{a}>b$ where $b<0$ is automatically true as soon as $a\ge0$.

*If $a\ge0$ and $b\ge0$, the inequality $a>b$ is equivalent to $a^2>b^2$.


So you have two cases to deal with.
Case 1
\begin{cases}
x^2-1\ge0\\
x<0
\end{cases}
which is equivalent to $x\le-1$
Case 2
\begin{cases}
x^2-1\ge0\\
x\ge0\\
(\sqrt{x^2-1})^2>x^2
\end{cases}
that has no solution, because the last inequality becomes $-1>0$, which is false.
A: For this to even make sense, $x$ must be in $(-\infty,-1]\cup[1,\infty)$. If it's in $[1,\infty)$, then
$$\sqrt{x^2-1}>x\iff x^2-1>x^2,$$
which is impossible. If, on the other hand, $x$ is in $(-\infty,-1]$, it's trivially true as the left side can't be negative.
A: The following solution is not the fastest but aims to explain how to select cases for $x$.
Firstly exclude the cases that $-1<x<1$ because in that case the term $x^2-1$ inside the square root would be negative, since $$x^2-1<0 \iff x^2<1 \overset{\sqrt{\cdot}}\iff |x|<1 \iff -1<x<1$$ Now, select cases 


*

*Let $x\ge 1$. Then you can divide both sides with $x$ without changing the direction of the inequality sign and you obtain $$\frac{\sqrt{x^2-1}}{x}>1 \iff \sqrt{\frac{x^2-1}{x^2}}>1\iff \sqrt{1-\frac{1}{x^2}}>1$$ But the root of the number is greater than $1$ if (and only if) the number is greater than $1$ so the last equation reduces to $$1-\frac{1}{x^2}>1 \iff -\frac{1}{x^2}>0$$ which is however impossible since $x^2>0$ (in particular for $x>1$ as we have assumed in this case).

*Let $x\le -1$. Then you can divide both sides with $x$ but now you have to change the direction of the inequality sign (since you divide with a negative number) and thus you obtain $$\frac{\sqrt{x^2-1}}{x}<1 \iff \sqrt{\frac{x^2-1}{x^2}}<1\iff \sqrt{1-\frac{1}{x^2}}<1$$ But the root of the number is less than $1$ if (and only if) the number is less than $1$ so the last equation reduces to $$1-\frac{1}{x^2}<1 \iff -\frac{1}{x^2}<0$$ which is always true since $x^2>0$ and therefore $$-x^2<0 \iff -\frac{1}{x^2}<0$$ So the inequality is true always under this cases, therefore the inequality holds for every $$x\le-1$$ and there are no other solutions as we saw with the previous cases. 

