Question

If $x$ satisfies $\left| {x - 1} \right| + \left| {x - 2} \right| + \left| {x - 3} \right| \geqslant 6$ then prove that all the numbers of $x$ which satisfy the above relation are given by $x \leqslant 0$ or $x \geqslant 4$.

Answer 1

To solve this problem we need to identify the change points of the equation.
From these points we will consider certain cases to find the range of $x$
The cases that fall into the range as well as satisfy the given condition
We will get the required answer.

Complete step-by-step solution:
Here the change points will be $x = 1,2,3$.
$(i)x \leqslant 1$ $(ii)1 < x \leqslant 2$ $(iii)2 < x \leqslant 3$ $(iv)x > 3$
We will find for each case if the value of x is possible such that it satisfies $\left| {x - 1} \right| + \left| {x - 2} \right| + \left| {x - 3} \right| \geqslant 6$
$(i)x \leqslant 1 \to (1)$
Here the given condition is$\left| {x - 1} \right| + \left| {x - 2} \right| + \left| {x - 3} \right| \geqslant 6$ can be written as $1 - x + 2 - x + 3 - x \geqslant 6$ since modulus of any number gives its absolute value.
$\Rightarrow 1 - x + 2 - x + 3 - x \geqslant 6$
We add the constant terms together and $x$ terms together.
$\Rightarrow 6 - 3x \geqslant 6$
Taking $6$ to right hand side as $- 6$
$\Rightarrow - 3x \geqslant 6 - 6$
Subtracting $6$ from $6$results in $0$
$\Rightarrow - 3x \geqslant 0$
Since there is a $-$ sign on the left-hand side, we will multiply both sides by $-$ to get the value of$x$.
In doing so, the inequality will change and result in
$\Rightarrow x \leqslant 0$
As we have considered in equation$\left( 1 \right)$, $x \leqslant 1$ means that the values of $x$ can be any value less than or equal to $1$ but not greater than$1$.
Hence $x \leqslant 0$ is possible when $x \leqslant 1$, this case is true.
$(ii)1 < x \leqslant 2 \to (2)$
Here $\left| {x - 1} \right| + \left| {x - 2} \right| + \left| {x - 3} \right| \geqslant 6$ can be written as $x - 1 + 2 - x + 3 - x \geqslant 6$
Since modulus of any number gives its absolute value.
$\Rightarrow x - 1 + 2 - x + 3 - x \geqslant 6$
We add the constant terms together and $x$ terms together.
$\Rightarrow 4 - x \geqslant 6$
Taking $4$ from left hand side to right hand side
$\Rightarrow - x \geqslant 6 - 4$
On subtracting the term and we get,
$\Rightarrow - x \geqslant 2$
Since there is a $-$ sign on the left-hand side, we will multiply both sides by $-$ to get the value of $x$.
In doing so, the inequality will change and result in
$\Rightarrow x \leqslant - 2$
As we have considered in equation $\left( 2 \right)$, $1 < x \leqslant 2$ means that the values of x can be any value greater than but not including $1$ up to$2$, and not greater than $2$.
Since $x \leqslant - 2$ cannot be possible when$1 < x \leqslant 2$, this case is not possible.
$(iii)2 < x \leqslant 3 \to (3)$
Here the given condition is $\left| {x - 1} \right| + \left| {x - 2} \right| + \left| {x - 3} \right| \geqslant 6$ can be written as $x - 1 + x - 2 + 3 - x \geqslant 6$ since modulus of any number gives its absolute value.
$\Rightarrow x - 1 + x - 2 + 3 - x \geqslant 6$
We add the constant terms together and $x$ terms together.
$\Rightarrow 2x - x \geqslant 6$
Let us subtract the term and we get,
$\Rightarrow x \geqslant 6$
As we have considered in equation $\left( 3 \right)$, $2 < x \leqslant 3$ means that the values of x can be any value greater than but not including $2$ up to $3$, and not greater than $3$.
Since $x \geqslant 6$ cannot be possible when $2 < x \leqslant 3$, this case is not possible.
$(iv)x > 3 \to (4)$
Here the given condition is $\left| {x - 1} \right| + \left| {x - 2} \right| + \left| {x - 3} \right| \geqslant 6$ can be written as $x - 1 + x - 2 + x - 3 \geqslant 6$
Since modulus of any number gives its absolute value.
$\Rightarrow x - 1 + x - 2 + x - 3 \geqslant 6$
We add the constant terms together and $x$ terms together.
$\Rightarrow 3x - 6 \geqslant 6$
We take 6 from left hand side to right hand side
$\Rightarrow 3x \geqslant 6 + 6$
On adding we get,
$\Rightarrow 3x \geqslant 12$
Let us divide $3$ on both side we get,
$\Rightarrow x \geqslant 4$
As we have considered in equation $\;\left( 4 \right)$, $x > 3$ means that the values of x can be any value greater than $3$ but not $3$ or any value less than$3$.
Since $x \geqslant 4$ is possible when$x > 3$, this case is true.
Now, from the results of the above $4$ cases, it is evident that when $x$ satisfies $\left| {x - 1} \right| + \left| {x - 2} \right| + \left| {x - 3} \right| \geqslant 6$ the value of $x$ will be given only when $x \leqslant 0$ or $x \geqslant 4$
Hence proved.

Note: In this type of question we need to be aware that the modulus of any terms will give its absolute value. That is the negative sign can be neglected.
For example, $\left| 5 \right| = \left| 5 \right|$ and $\left| { - 25} \right| = \left| {25} \right|$
Here some of the properties of the modulus value as follow:
$\left| x \right| + \left| y \right| \geqslant \left| {x + y} \right|$
$\left| 0 \right| = 0$
$\left| x \right| \geqslant 0$
The next thing we need to be careful when we deal with inequalities; If we have to multiply or divide both sides of an inequality by a negative number, we will reverse the direction of the given inequality sign.