GS. 1 Let $\alpha$ and $\beta$ be the real roots of $y^{2}-6 y+5=0$. Let $m$ be the minimum value of $|x-\alpha|+|x-\beta|$ over all real values of $x$. Find the value of $m$.
Reference: 1994 HG1, 2001 HG9, 2004 FG4.2, 2008 HI8, 2008 FI1.3, 2010 HG6, 2012 FG2.3
$\alpha=1, \beta=5$
If $x4$
If $1 \leq x \leq 5,|x-\alpha|+|x-\beta|=x-1+5-x=4$
If $x>5,|x-\alpha|+|x-\beta|=x-1+x-5=2 x-6>4$
$m=$ min. of $|x-\alpha|+|x-\beta|=4$
Method 2 Using the triangle inequality: $|a|+|b| \geq|a+b|$
$$
|x-\alpha|+|x-\beta| \geq|x-1+5-x|=4 \Rightarrow m=4
$$

Remark: there is a typing mistake in the English version. ... minimum value $a$ of ...