$\displaystyle \forall x,y\in\mathbb{R}$

$\displaystyle \text{max}\{x,y\}=\frac{1}{2}\left(x+y+|x-y|\right)$

I know

$\displaystyle \text{max}\{|x|-|y|,|y|-|x|\}\leq |x-y|$

Results 1 to 4 of 4

- May 24th 2011, 11:59 AM #1

- Joined
- Mar 2010
- From
- Florida
- Posts
- 3,093
- Thanks
- 10

- May 24th 2011, 12:11 PM #2

- May 24th 2011, 12:12 PM #3

- Joined
- Mar 2010
- From
- Florida
- Posts
- 3,093
- Thanks
- 10

- May 24th 2011, 05:53 PM #4

- Joined
- May 2011
- Posts
- 10

Usually one would consider non-overlapping cases, but I guess the overlapping cases here would work.

Case 1: Suppose x $\displaystyle \leqslant $ y. What does this imply? If you don't recall what this implies I'll say that addition on**R**satisfy commutation, monotonicity, an equality for inverses, and an equality for neutrals. In other words, for all x, y, z, c on**R**the following properties for + hold:

If x$\displaystyle \leqslant $y, (x+c)$\displaystyle \leqslant $(y+c) (monotonicity)

(x+y)=(y+x) (commutation)

(x+-(x))=0 (inverse)

x+0=x (neutrality)

Since you started with x$\displaystyle \leqslant $y, do you see which property you want to use here first? Do you see what element (function of an element, depending on how you look at it) of**R**you want to select for "c"? If you do, I think you'll infer what I've hinted at. Then check your definition of absolute value. So in this case, abs(x-y) must equal... what? Then replace what abs(x-y) must equal in this case with abs(x-y) in the maximum formula.

Case 2: Suppose x>y. Now I'll point out that for all x, y, c in**R**,

If x>y, then (x+c)>(y+c). Now using a similar technique, along with your definition of absolute value you can show something about the absolute value in this case.

Then you might to want to indicate the cases exhaust**R**and the result consequently follows.

If the above doesn't give you enough clues, I'll happily try and spell out details here.