[SOLVED] Absolute values

**Bruno J.** · May 13th 2010, 11:56 AM

My own problem! Enjoy.

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

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

**simplependulum** · May 13th 2010, 09:39 PM

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 $a ,b ,c$ ?

**Bruno J.** · May 13th 2010, 09:41 PM

Yes!

**Drexel28** · May 13th 2010, 09:44 PM

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.

**simplependulum** · May 13th 2010, 09:47 PM

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 .

**Bruno J.** · May 13th 2010, 10:18 PM

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!

**undefined** · May 13th 2010, 11:42 PM

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.

**Unbeatable0** · May 14th 2010, 12:19 AM

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.

**Bruno J.** · May 14th 2010, 09:10 AM

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!

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