<s>[INST] Spivak Calculus chapter 1 problem 13 proof critique 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. 
 [/INST] 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. 
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