# 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.

• It may help to note that $|x| = \max(x,-x)$. – copper.hat Jun 26 '13 at 3:09
• I would start by convincing yourself that the proof is meaningful with several concrete examples. Set x=1, y=-3, for example, and try to get an intuition that it's true (before you try to prove why it's true). Then many of the comments above should be helpful. – Dave B. Jun 26 '13 at 4:23
• Just do a case-by-case analysis: Evaluate both sides once for the case $x\leq y$ and once for $y\leq x$. – Andy Brandi Jun 26 '13 at 6:23
• possible duplicate of Proof the maximum function $\max(x,y) = \frac {x +y +|x-y|} {2}$ – Analysis Mar 7 '14 at 13:43
• @TuckerRapu: IIRC, here and in a couple of other places you've marked an older question as a duplicate of a newer question. – Andrew D. Hwang Mar 7 '14 at 14:52

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}

• +1 it would be nice, if you noted the $\text{min}(x,y)$ for the OP as well. :) – mrs Jun 26 '13 at 5:00
• @BabakS.: Thanks! The $\min$ result follows immediately from $\max(x,y)+\min(x,y) = x+y$. – copper.hat Jun 26 '13 at 5:40
• Great! This should be accepted! – vonPetrushev Jun 26 '13 at 12:57

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.

• 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. – Ennar Oct 3 '15 at 19:23

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 $w\lt 0$.

• +1 This is the most straightforward one here. Also @user71925 it should also be noted that $\min\{x,y\} = \frac{1}{2}(x+y-|x-y|)$. – chs21259 May 20 '14 at 18:40

$$\max\{x,y\} =\frac{x+y+|x-y|}{2}$$ $$=> 2.\max\{x,y\} =x+y+|x-y|$$ 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 $x>y$

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.

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.