Proving $\max$ of $a, b$. How do I prove that $$\max{\{a, b\}} = \frac{a + b + \left | a - b \right |}{2}$$
I have no idea how to even start the proof, any idea / intuition that can get me started is greatly appreciated.
Thanks for reading my post.
 A: *

*If $a\geq b$, then $|a-b| = a-b$ and $\max\{a,b\} = a$. Now, simplify the expression you want to prove.

*If $b\geq a$, then I think you can finish the sentence.

A: A tricky way is to notice that
\begin{align*}
\max(a,b) + \min(a,b) &= a+b \\ 
\max(a,b) - \min(a,b) &= |a-b| \\
\end{align*}
Add these two equations together and we have 
$$2 \max(a,b) = a+b + |a-b|$$
and your result follows. 
A: The intuition behind this is the following. Write the RHS as $$\frac{a+b}{2}+\frac{|a-b|}{2}$$ Now the first term $\dfrac{a+b}{2}$ gets you exactly at the midpoint between $a$ and $b$. For example if the numbers $a,b$ are $5,10$ then this gets you to $7.5$. So, to reach the maximum of the two you need to add the half of their distance, or in symbols $$+\frac{d(a,b)}{2}$$ In our example this would be $5/2=2.5$ and indeed $7.5+2.5=10$ is the maximum of the two numbers. Now, you need to know that the distance $d(a,b)$ of two numbers $(a,b)$ is given by the absolute value of their difference, that is $d(a,b)=|a-b|$ which gives you the result. 
Unfortunately this expression does not admit a generalization for $3$ or more numbers.
