# [SOLVED] Absolute values

#### Bruno J.

MHF Hall of Honor
My own problem! Enjoy. (Smirk)

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

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

#### simplependulum

MHF Hall of Honor
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 $$\displaystyle a ,b ,c$$ ?

#### Bruno J.

MHF Hall of Honor
Yes!

#### Drexel28

MHF Hall of Honor
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.

#### simplependulum

MHF Hall of Honor
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 .

#### Bruno J.

MHF Hall of Honor
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)

#### undefined

MHF Hall of Honor
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.

#### Unbeatable0

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.

• Chris L T521, Bruno J. and undefined

#### Bruno J.

MHF Hall of Honor
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)