My own problem! Enjoy.
Show that for $\displaystyle a,b,c \in \mathbb{R}$, one has
$\displaystyle |a-b|+c+\Big | a+b+|a-b|-2c\Big | = |a-c|+b+\Big | a+c+|a-c|-2b\Big |$
Is there a clever way to do this or just brute stuff? If you created it you probably worked backwards from something and so have an "ideal" solution which retraces your steps. I would guess the easiest way to do this would be to have that $\displaystyle f(a,b,c)$ be the LHS and then we are trying to show that $\displaystyle f(a,b,c)=f(a,c,b)$ and to start squaring stuff.
This approach is basically brute force...
There are 3! ways to permute {a,b,c}, leading to 6 cases:
case $\displaystyle a \leq b \leq c$
$\displaystyle |a-b|+c+\Big | a+b+|a-b|-2c\Big | =?\ |a-c|+b+\Big | a+c+|a-c|-2b\Big |$
$\displaystyle b-a+c+ | a+b+b-a-2c | =?\ c-a+b+| a+c+c-a-2b |$
$\displaystyle b-a+c+ | 2b-2c | =?\ c-a+b+| 2c-2b |$
$\displaystyle b-a+c+ 2(c-b) = c-a+b+2(c-b)$
I won't list the other cases because they all work out similarly, unless I'm missing something.
We have $\displaystyle \max\{a,b\} = \frac{a+b+|a-b|}{2}$. Thus,
$\displaystyle \max\{a,b,c\} = \max\{a,\max\{b,c\}\} = \frac{1}{4}(2a+b+c+|b-c|+\Big|b+c+|b-c|-2a\Big|)$
We can switch $\displaystyle a$, $\displaystyle b$ and $\displaystyle c$ in the above identity and still get $\displaystyle \max\{a,b,c\}$. So:
$\displaystyle \text{RHS when } a \leftrightarrow c = \text{RHS when } a \leftrightarrow b$
$\displaystyle
\frac{1}{4}(2c+a+b+|a-b|+\Big|a+b+|a-b|-2c\Big|) =
\frac{1}{4}(2b+a+c+|a-c|+\Big|a+c+|a-c|-2b\Big|)
$
Multiplying by $\displaystyle 4$ and subtracting $\displaystyle a+b+c$ from both sides yields the result.