1. ## [SOLVED] Absolute values

My own problem! Enjoy.

Show that for $a,b,c \in \mathbb{R}$, one has

$|a-b|+c+\Big | a+b+|a-b|-2c\Big | = |a-c|+b+\Big | a+c+|a-c|-2b\Big |$

2. Originally Posted by Bruno J.
My own problem! Enjoy.

Show that for $a,b,c \in \mathbb{R}$, one has

$|a-b|+c+\Big | a+b+|a-b|-2c\Big | = |a-c|+b+\Big | a+c+|a-c|-2b\Big |$

We need to show that it is an identity which holds for all real numbers $a ,b ,c$ ?

3. Yes!

4. Originally Posted by Bruno J.
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 $f(a,b,c)$ be the LHS and then we are trying to show that $f(a,b,c)=f(a,c,b)$ and to start squaring stuff.

5. Originally Posted by Drexel28
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 $f(a,b,c)$ be the LHS and then we are trying to show that $f(a,b,c)=f(a,c,b)$ and to start squaring stuff.

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

6. Originally Posted by Drexel28
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 $f(a,b,c)$ be the LHS and then we are trying to show that $f(a,b,c)=f(a,c,b)$ and to start squaring stuff.
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!

7. Originally Posted by Bruno J.
My own problem! Enjoy.

Show that for $a,b,c \in \mathbb{R}$, one has

$|a-b|+c+\Big | a+b+|a-b|-2c\Big | = |a-c|+b+\Big | a+c+|a-c|-2b\Big |$
This approach is basically brute force...

There are 3! ways to permute {a,b,c}, leading to 6 cases:

case $a \leq b \leq c$

$|a-b|+c+\Big | a+b+|a-b|-2c\Big | =?\ |a-c|+b+\Big | a+c+|a-c|-2b\Big |$

$b-a+c+ | a+b+b-a-2c | =?\ c-a+b+| a+c+c-a-2b |$

$b-a+c+ | 2b-2c | =?\ c-a+b+| 2c-2b |$

$b-a+c+ 2(c-b) = c-a+b+2(c-b)$

I won't list the other cases because they all work out similarly, unless I'm missing something.

8. Originally Posted by Bruno J.
My own problem! Enjoy.

Show that for $a,b,c \in \mathbb{R}$, one has

$|a-b|+c+\Big | a+b+|a-b|-2c\Big | = |a-c|+b+\Big | a+c+|a-c|-2b\Big |$
We have $\max\{a,b\} = \frac{a+b+|a-b|}{2}$. Thus,

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

We can switch $a$, $b$ and $c$ in the above identity and still get $\max\{a,b,c\}$. So:

$\text{RHS when } a \leftrightarrow c = \text{RHS when } a \leftrightarrow b$

$
\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 $4$ and subtracting $a+b+c$ from both sides yields the result.

9. Originally Posted by Unbeatable0
We have $\max\{a,b\} = \frac{a+b+|a-b|}{2}$. Thus,

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

We can switch $a$, $b$ and $c$ in the above identity and still get $\max\{a,b,c\}$. So:

$\text{RHS when } a \leftrightarrow c = \text{RHS when } a \leftrightarrow b$

$
\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 $4$ and subtracting $a+b+c$ from both sides yields the result.
Good job!