Spivak Calculus chapter 1 problem 13 proof critique

Question

Question paraphrased: Prove that the maximum between two numbers $x$ and $y$ is given by:
$$\max(x,y)=\frac{x+y+|y-x|}{2}$$ Proof: Let $x$ and $y$ be two arbitrary numbers. Then, one and only one holds true: $x \ge y$ or $x \le y$
If $x \ge y$, $x=x+(y-y)=(x-y)+y=|x-y|+y$
Also, $$x=\frac {x+x}{2}=\frac{|x-y|+y+x}{2}=\frac{|x-y|+x+y}{2}$$

If $y \ge x$, $y=y+(x-x)=(y-x)+y=|y-x|+y$
Also, $$y=\frac {y+y}{2}=\frac{|y-x|+x+y}{2}=\frac{|x-y|+x+y}{2}$$
Thus, $$\max(x,y)=\frac{x+y+|y-x|}{2}$$
How can I improve this proof? I am not very convinced with $y=\frac {y+y}{2}$ and $x=\frac {x+x}{2}$ technique because it feels as if I have done this to satisfy the answer.

Answer 1

Geometrically, suppose that $p$ and $q$ lie on real $x$ axis then,
max$(p, q)$= middle point + half the distance between $(p, 0)$ and $(q, 0)$
$= \frac{p+q} {2} + \frac{|p-q|} {2}$

Answer 2

A quicker way to the proof would be to show LHS=RHS. If $x\geq y,$ then LHS$=x$ and $|y-x|=-(y-x)=x-y$ which gives $$\text{RHS }=\frac{x+y+(x-y)}{2}=\frac{x+x}{2}=x.$$

The case $y> x$ is similar.

Answer 3

Your proof is ok.

Rephrasing:

1) Let $y\ge x$:

Then $\max(x,y)=y$ and

$\dfrac{y+x+|y-x|}{2}=2y/2=y$, since $|y-x|=y-x$.

2) Let $y<x$ :

Then $\max(x,y)=x$ and

$\dfrac{y+x+|y-x|}{2}=x$ since $|y-x|=x-y$.