Proving $|a-1|+|a-2|+|a-3| \ge 2$

Question

I need to prove the following sentence for $a\in\mathbb{R}$:

$$ |a-1|+|a-2|+|a-3| \ge 2$$

Breaking the equation into cases it does work, i.e. for $a\le 1$:

$$-a+1-a+2-a+3\ge 2$$ $$-3a \ge -4$$ $$a \le 4/3$$ Which is always true since $a \le 1$, but writing all the cases like that doesn't really seem like a sufficient proof, or a very smart way of doing it. Is there a better way to prove it?

Answer 1

Apply the triangle inequality $|x|+|y|\geq |x+y|$ three times: \begin{align*} |a-1|+|a-2|&=|a-1|+|2-a|\geq 1\\ |a-2|+|a-3|&=|a-2|+|3-a|\geq 1\\ |a-3|+|a-1|&=|3-a|+|a-1|\geq 2. \end{align*} Summing these and dividing by $2$ gives the desired result.

Answer 2

You can use the triangle inequality of the absolute value: $$ |a\pm b| \leq |a|+|b|$$ Now $2 = |2| = |(a-1)-(a-3)| \leq |a-1|+|a-3|$ so in fact $$|a-1|+|a-2|+|a-3| \geq 2+|a-2| \geq 2$$ Notice that equality is achieved when $a = 2$.

Answer 3

can't we argue this geometrically. $$|a-1| + |a-2| + |a-3|$$ is the sum of the distances of the point $a$ on the number line from the three points $1, 2$ and $3.$ if the point $a$ is between $1$ and $3,$ then the distances from $1$ and $3$ themselves add up to $2,$ therefore the three distances must be $\ge 2.$ if you take a point outside then either of the distances from $1$ or $3$ must be $\ge 2.$ that takes care of all the possibilities.

Answer 4

The triangle inequality doesn't seem to be necessary: $$\underbrace{|a-1|}_{\ge a-1}+\underbrace{|a-2|}_{\ge0}+\underbrace{|a-3|}_{\ge 3-a}\ge(a-1)+0+(3-a)=2$$

Answer 5

It is known that the number $a$ minimizing $\sum_i |a-x_i|$ is the median of the $a_i$; a proof of this already appears on the site (if you can find it!). In our case, this means $$ |a-1| + |a-2| + |a-3| \geq |2-1| + |2-2| + |2-3| = 2. $$