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