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 $x>1$.