Solve $\vert x-2\vert+2\vert x-4\vert\leq \vert x+1\vert$

Question

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

Answer 1

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.

Answer 2

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 $\left(\frac{11}{2},\infty\right)$.

Answer 3

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: $\dfrac{5}{2}\le x\le\dfrac{11}{2}$.

Answer 4

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 $x\in[2.5,5.5]$.