Looking at the solution of Trench´s Introduction to Real Analysis exercises, I am struggling with this:
Write the following expression in equivalent form not involving absolute values:
$a + b + 2c + |a - b| + |a + b - 2c + |a - b||$
it uses the following solution
$(a)$ $a + b + |a - b| = 2\max(a, b)$
Let $\alpha=a + b + 2c + |a - b| + |a + b - 2c + |a - b||$
From $(a)$, $\alpha = 2[\max(a,b)+c+|\max(a,b)-c|] = df\beta$. From $(a)$ with $a$ and $b$ replaced by $max(a,b)$ and $c$, $\beta = 4\max(\max(a,b),c) = 4\max(a,b,c)$
what I don´t quite grasp is the $df\beta$ part that´s then changed to $\beta$, how do they come up with that?
It uses part $(a)$, so for $2(\max(a,b)+c+|\max(a,b)-c|)$ Let $\max(a,b)=\delta$, then we have:
$\max(a,b)+c+|\max(a,b)-c|=\delta+c+|\delta-c|=2\max(\delta,c)$ (from $(a)$)
Thus:
$2(\max(a,b)+c+|\max(a,b)-c|)=2(2\max(\delta,c))=4\max(\delta,c)=4\max(\max(a,b),c)$
So in the end:
$\max(a,b)+c+|\max(a,b)-c|=4\max(\max(a,b),c)$.