after playing around with absolute value for a little, i noticed that $|a + b| + |a - b|$ always appeared to be the larger of $a$ and $b$ multiplied by $2$, whenever both $a$ and $b$ were positive.
$$ |a + b| + |a - b| = \begin{cases} 2a, & \text{if $a \ge b$} \\ 2b, & \text{if $a < b$} \end{cases} $$
can this be proven to always be the case, and if so, how do i prove it?
As this covers all cases for real $a,b$, we have $$ a+b+|a-b|=2\max\{a,b\}$$ for all $a,b\in\Bbb R$.
Answer posted to provide industrial strength approach for new students.
To prove:
$|a + b| + |a - b| = 2\max(a,b).$
Either $a \geq b,$ or $a < b.$
Further, either $(a+b) \geq 0$ or $(a+b) < 0.$
Case 1:
$a \geq b$ and $(a+b) \geq 0.$
Then $|a + b| + |a - b| = (a+b) + (a-b) = 2a = 2\max(a,b).$
The assertion is always true in this case.
Case 2:
$a \geq b$ and $(a+b) < 0.$
Then $|a + b| + |a - b| = -(a + b) + (a - b) = -2b.$
$2a = 2\max(a,b).$
The assertion is not always true in this case.
Case 3:
$a < b$ and $(a+b) \geq 0.$
Then $|a + b| + |a - b| = (a+b) + (b-a) = 2b = 2\max(a,b).$
The assertion is always true in this case.
Case 4:
$a < b$ and $(a+b) < 0.$
Then $|a + b| + |a - b| = -(a+b) + (b - a) = -2a.$
$2b = 2\max(a,b).$
The assertion is not always true in this case.
