# [SOLVED] Absolute values

• May 13th 2010, 11:56 AM
Bruno J.
[SOLVED] Absolute values
My own problem! Enjoy. (Smirk)

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

$\displaystyle |a-b|+c+\Big | a+b+|a-b|-2c\Big | = |a-c|+b+\Big | a+c+|a-c|-2b\Big |$
• May 13th 2010, 09:39 PM
simplependulum
Quote:

Originally Posted by Bruno J.
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 $\displaystyle a ,b ,c$ ?
• May 13th 2010, 09:41 PM
Bruno J.
Yes!
• May 13th 2010, 09:44 PM
Drexel28
Quote:

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 $\displaystyle f(a,b,c)$ 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.
• May 13th 2010, 09:47 PM
simplependulum
Quote:

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 $\displaystyle f(a,b,c)$ 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 .
• May 13th 2010, 10:18 PM
Bruno J.
Quote:

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 $\displaystyle f(a,b,c)$ 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.

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)
• May 13th 2010, 11:42 PM
undefined
Quote:

Originally Posted by Bruno J.
My own problem! Enjoy. (Smirk)

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

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

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.
• May 14th 2010, 12:19 AM
Unbeatable0
Quote:

Originally Posted by Bruno J.
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.
• May 14th 2010, 09:10 AM
Bruno J.
Quote:

Originally Posted by Unbeatable0
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.

Good job! (Clapping)