<s>[INST] 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
 [/INST] 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 $\left(\frac{11}{2},\infty\right)$.
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