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 need to show that it is an identity which holds for all real numbers \(\displaystyle a ,b ,c \) ?

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.Yes!

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.

Clever way ? Is it useful to draw a geometric figure ? Haha , sounds quite silly .

There is a very clever way! A small hint would give it all away... You don't even really have to do any calculations, even less draw a diagram! (Nod)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...My own problem! Enjoy. (Smirk)

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

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

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,My own problem! Enjoy. (Smirk)

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

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

\(\displaystyle \max\{a,b,c\} = \max\{a,\max\{b,c\}\} = \frac{1}{4}(2a+b+c+|b-c|+\Big|b+c+|b-c|-2a\Big|)\)

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.

Good job! (Clapping)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.