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
 A: There are three points of interest: $-1, 2, 4$.  These divide the real line into four segments, so the problem is solved with four cases:
Case 1: $x\ge 4$.  Now we have $(x-2)+2(x-4)\le (x+1)$, which we rearrange as $2x\le 11$, or $x\le 5.5$.  Since $x\ge 4$, this gives us a range of solutions $4\le x\le 5.5$.
Case 2: $2\le x\le 4$. Now we have $x-2+2(4-x)\le (x+1)$, which we rearrange as $x\ge 2.5$.  This gives us the interval $2.5\le x\le 4$.
I leave the other two cases for you to solve.
A: 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 $\left(\frac{11}{2},\infty\right)$.
A: 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: $\dfrac{5}{2}\le x\le\dfrac{11}{2}$. 
A: A (mostly) geometric answer (meaning it uses the triangle inequality):
$$|x-2|+|2x-8|\leq|x+1|\implies|(2x-8)-(x-2)|\leq|x+1|\implies|x-6|\leq|x+1|$$
So $x$ has to be at least as close to $6$ as to $-1$. So $x\geq2.5$ (their average).
But also
$$
\begin{align}
|x-2|+|2x-8|\leq|x+1|&\implies|3x-10|\leq|x+1|\\
&\implies9x^2-60x+100\leq x^2+2x+1\\
&\implies8x^2-62x+99\leq0\\
&\implies(4x-9)(2x-11)\leq0\\
&\implies x\in[2.25,5.5]
\end{align}$$
Taken together, it is necessary that $x\in[2.5,5.5]$. But this is also sufficient. If $x\in[2.5,5.5]$, then the inequality reduces to $$x-2+2|x-4|\leq x+1$$ $$\Longleftrightarrow |x-4|\leq \frac32$$ which is true for $x\in[2.5,5.5]$.
