My own problem! Enjoy. (Smirk)

Show that for \(\displaystyle a,b,c \in \mathbb{R}\), one has

\(\displaystyle |a-b|+c+\Big | a+b+|a-b|-2c\Big | = |a-c|+b+\Big | a+c+|a-c|-2b\Big |\)

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.