Solving inequation with two absoulte values I need to solve the following inequation:
$$
|x| \cdot |x-1|-1>-x\\
$$
I cant get the correct result.
I tried to solve it like this:
$$
|x| \cdot |x-1|-1>-x
$$
I know that I can write $|x \cdot y|=|x| \cdot |y|$, so:
$$
\begin{eqnarray}
|x(x-1)|-1&>&-x\\
|x^2-x|-1&>&-x \qquad \text{for }\; x < 0 \\
-(x^2-x)-1&>&-x\\
-x^2+x-1&>&-x\\
-x^2+x+x-1&>&0\\
-x^2+2x-1&>&0
\end{eqnarray}
$$
but I think that this is incorrect, because when I solve this inequality I get that $x<1$ and $x>1$, and that doesn't make a lot of sense
Any suggestion is helpful. Than you very much!!!

You should distinguish three cases:


*

*when $x<0$, $|x|=-x$ and $|x-1|=1-x$;

*when $0\le x\le 1$, $|x|=x$ and $|x-1|=1-x$;

*when $x>1$, $|x|=x$ and $|x-1|=x-1$.
Thus you have to solve
\begin{align}
&\begin{cases}
(-x)(1-x)-1>-x\\
x<0
\end{cases}\\
&\begin{cases}
x(1-x)-1>-x\\
0\le x\le 1
\end{cases}\\
&\begin{cases}
x(x-1)-1>-x\\
x>1
\end{cases}
\end{align}
and put the solutions sets together.

Alternative solution, which, however, could bring to more complicated computations in other cases.
Write the inequation as
$$
|x^2-x|>1-x
$$
We can see that any value of $x$ such that $1-x<0$ is a solution. So $x>1$ is always a solution.
In case $1-x>0$, we can square, so we get
$$
\begin{cases}
x^2(1-x)^2>(x-1)^2\\
1-x>0
\end{cases}
$$
This simplifies to
$$
\begin{cases}
x^2>1\\
x<1
\end{cases}
$$
which is satisfied for $x<-1$.
For $x=1$ the inequality doesn't hold
Putting together the two solution sets, we conclude that the inequality is satisfied for $x<-1$ or $x>1$.
