#### [SOLVED] Show that the $\max{ \{ x,y \} }= \frac{x+y+|x-y|}{2}$.

Show that the $\max{ \{ x,y \} }= \dfrac{x+y+|x-y|}{2}$.

I do not understand how to go about completing this problem or even where to start.

#### @copper.hat 2013-06-26 03:12:43

Here is another way of looking at it:

We have $|x| = \max(x,-x)$. Also, $\max(a,b)+c = \max(a+c,b+c)$, and if $c \geq 0$, then $c \max(a,b) = \max(ac,bc)$.

Hence \begin{eqnarray} \frac{1}{2}(x+y+|x-y|) &=& \frac{1}{2}(x+y+\max(x-y,y-x)) \\ &=& \frac{1}{2}(\max(x-y+x+y,y-x+x+y)) \\ &=& \frac{1}{2}(\max(2x,2y)) \\ &=& \max(x,y) \end{eqnarray}

#### @mrs 2013-06-26 05:00:54

+1 it would be nice, if you noted the $\text{min}(x,y)$ for the OP as well. :)

#### @copper.hat 2013-06-26 05:40:24

@BabakS.: Thanks! The $\min$ result follows immediately from $\max(x,y)+\min(x,y) = x+y$.

#### @vonPetrushev 2013-06-26 12:57:57

Great! This should be accepted!

#### @ntomlin1996 2014-05-20 18:26:18

Without loss of generality, let $y=x+k$ for some nonnegative number $k$. Then,

$$\frac{x+(x+k)+|x-(x+k)|}{2} = \frac{2x+2k}{2} = x+k = y$$

which is equal to $\max(x,y)$ by the assumption.

#### @Kevin Pardede 2013-06-26 07:30:11

$$\max\{x,y\} =\frac{x+y+|x-y|}{2}$$ $$=> 2.\max\{x,y\} =x+y+|x-y|$$ there are two possible situation :
1. $y>x$, i.e $\max\{x,y\}=y$ then $y-x=|x-y|$, this equation is true because we assume that
$y>x$
2. $x>y$, or $\max\{x,y\}=x$, then $x-y=|x-y|$, which is true if $x>y$

#### @Uthsav Chitra 2013-06-26 03:12:20

This probably isn't as rigorous as it should be, but I think it's intuitive enough.

Hmm... We don't know which of $x$ or $y$ is bigger, but we do know one thing: their average. If we call the average $z$, then $z=\frac{x+y}{2}$. Now, the distance between $x$ and $y$ is $|x-y|$, so the distance from $z$ to both $x$ and $y$ is $\frac{|x-y|}{2}$.

So if we imagine a number line, the distance from $0$ to $z$ is $\frac{x+y}{2}$, and the distance from $z$ to max(x, y) is $\frac{|x-y|}{2}$. Thus, the total distance from $0$ to max(x, y) is $\frac{|x-y|}{2}$ + $\frac{x+y}{2}$, as desired.

#### @Ennar 2015-10-03 19:23:21

I like this answer. If not as rigorous, this is exactly how one would come up with the expression in the first place. Also, one can easily come up with similar expression for $\operatorname{min}$ in this manner.

#### @James 2013-06-26 03:16:39

Conceptually, focus on $|x-y|$ as the absolute difference of the two numbers.

Without loss of generality, assume $x > y$. Then $y + |x - y| = x$. This can be understood as representing that if we add the difference between two numbers to the smaller, we get the larger.

The specific equation will naturally fall out with this observation.

#### @André Nicolas 2013-06-26 03:07:55

Hint: If $x\ge y$ then $|x-y|=x-y$. If $x\lt y$ then $|x-y|=-(x-y)=y-x$.

We have used the fact that in general $|w|=w$ if $w\ge 0$ and $|w|=-w$ if $w\lt 0$.

#### @chs21259 2014-05-20 18:40:58

+1 This is the most straightforward one here. Also @user71925 it should also be noted that $\min\{x,y\} = \frac{1}{2}(x+y-|x-y|)$.