Solve inequality with absolute values $|3-5x|-|2x+1|\ge|x-1|$

Question

How can I solve this: $$|3-5x|-|2x+1|\ge|x-1|$$ $x=3/5,-1/2,1$ is when the absolutes is $0$. I received $x=3/2$ when the equality holds. But for every $x$ the equation $|3-5x|-|2x+1|>|x-1|$. What I am doing wrong?

Answer 1

The function $f(x):=|3-5x|-|2x+1|-|x-1|$ is piecewise linear and the "corners" are located where one of the arguments goes to zero, i.e. $x=-\dfrac12,\dfrac35,1$.

Evaluating the function at these points (using the appropriate sign in $|t|=\pm t$), we get

$$f\left(-\frac12\right)=4,f\left(\frac35\right)=-\frac{13}5,f\left(1\right)=-1$$

so that the function has two roots, one in $\left(-\dfrac12,\dfrac35\right)$ and one in $(1,\infty)$.

Now, expressing the function in these ranges, we get

$$\left(-\dfrac12,\dfrac35\right)\to3-5x-2x-1\ge1-x\implies6x\le1,$$ $$\left(1,\infty\right)\to5x-3-2x-1\ge x-1\implies2x\ge3.$$