Solve $\vert x-2\vert+2\vert x-4\vert\leq \vert x+1\vert$ I was helping someone with abolute values and inequalities and found this question.
What is the easiest way to solve this?
The only thing I thought of is to add the L.H.S and graph it with the R.H.S to answer the questoin is there simpler way to deal with this?
Thank you

You have:
$$
x-2\ge 0 \iff x\ge 2
$$
$$
x-4\ge 0 \iff x\ge 4
$$
$$
x+1\ge 0 \iff x\ge -1
$$
so we can split the inequality $|x-2|+2|x-4|-|x+1|\le 0$ in four systems:
$$
\begin{cases}
x<-1\\
2-x+2(4-x)+x+1\le 0
\end{cases}
$$
$$
\begin{cases}
-1\le x<2\\
2-x+2(4-x)-x-1\le 0
\end{cases}
$$
$$
\begin{cases}
2\le x<4\\
x-2+2(4-x)-x-1\le 0
\end{cases}
$$
$$
\begin{cases}
4\le x\\
x-2+2(x-4)-x-1\le 0
\end{cases}
$$
and the solution of the inequality is the union of the solutions of these systems.
With a bit of algebra you can see that the first two systems have no solutions, and the solutions of the other two are: $ \dfrac{5}{2}\le x<4$ and $4\le x \le \dfrac{11}{2}$, so the final solution is the union: $\dfrac{5}{2}\le x\le\dfrac{11}{2}$. 
