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

An alternate non case-wise approach that involves more calculation with larger numbers:
Look for equality first. Squaring preserves the equality (although it may introduce extraneous solutions which can be ruled out at the end).
$$
\begin{align}
(x-2)^2+4|(x-2)(x-4)|+4(x-4)^2
&=(x+1)^2\\
x^2-4x+4+4|(x-2)(x-4)|+4x^2-32x+64
&= x^2+2x+1\\
4|(x-2)(x-4)|
&=-4x^2+38x-67\\
16x^4-192x^3+832x^2-1536x+1024
&=16x^4-304x^3+1980x^2-5092x+4489\\
112x^3-1148x^2+3556x-3465
&=0\\
16x^3-164x^2+508x-495
&=0\\
(4x-9)(2x-5)(2x-11)
&=0\\
\end{align}
$$
where the final factorization uses the rational root theorem. Checking, $\frac{9}{4}$ does not give equality, but both $\frac{5}{2}$ and $\frac{11}{2}$ do. Now check the direction of inequality on $\left(-\infty,\frac52\right)$, $\left(\frac52,\frac{11}{2}\right)$, and $\left(\frac{11}{2},\infty\right)$.
