Proof that $\frac{(x+y)-abs(x-y)}{2}$ equivalent to $\min(x,y)$ I plotted the two functions $\frac{(x+y)-abs(x-y)}{2}$ and $min(x,y)$ in the range $[-1, 1]$ and they look the same. The both $min$ and $abs$ functions are defined as expected.
$abs(x)=\begin{cases}&x,&0<x,\\-&x,&\text{else.}\end{cases}$
$\min(x,y)=\begin{cases}x,&x<y,\\y,&\text{else.}\end{cases}$

The $min$ function is quite easy to imagine. For the other function, I think of it as the average of $x$ and $y$ minus their half distance. Anyhow, I haven't an imagination of both functions being equal yet. Could you provide a proof and the idea behind it to me?
 A: The proof is simple -- just consider the two possible cases:


*

*If $x\geq y$, the absolute value is equal to plain $(x-y)$ and $\min(x,y)=y$ and left-hand side is equal to the right-hand side.

*If $x<y$, the absolute value is equal to $(y-x)$ and $\min(x,y)=x$. Again, LHS = RHS.


Since these are the only possible cases, the proof is complete.
The intuition behind it is exactly as you described it -- $\frac{x+y}{2}$ is exactly halfway between $x$ and $y$, so if we move this half-distance "down" (= towards smaller numbers), we'll reach the smaller of the two numbers. If we added $|x-y|$ instead of subtracting it, we'd get $\max(x,y)$.
A: Note that $-|x| = \min\{-x, x\}$. 
We want to relate $\min\{x, y\}$ to $|x|$. We'll use the fact that for any $a$ we have $$\min\{x, y\} = \min\{x - a, y - a\} + a.$$ If $x - a$ and $y - a$ were negatives of one another, we would be able to replace $\min\{x - a, y - a\}$ by $-|x - a|$ (which would also be equal to $-|y-a|$). So the question becomes, can we choose $a$ such that $x - a = -(y - a)$? The answer is yes. Rearranging this equation we get $a = \frac{1}{2}(x+y)$. Therefore we have 
\begin{align*}
\min\{x, y\} &= \min\{x - a, y -a\} + a\\ 
&= \min\left\{x - \frac{1}{2}(x+y), y - \frac{1}{2}(x+y)\right\} + \frac{1}{2}(x+y)\\
&= \min\left\{\frac{1}{2}(y-x), \frac{1}{2}(x - y)\right\} + \frac{1}{2}(x+y)\\
&= \min\left\{-\frac{1}{2}(x-y), \frac{1}{2}(x-y)\right\} + \frac{1}{2}(x+y)\\
&= -\left|\frac{1}{2}(x-y)\right| + \frac{1}{2}(x+y)\\
&= \frac{1}{2}(x+y-|x+y|).
\end{align*}
If you think about $x$ and $y$ on the numberline, what we've done is found the midpoint of $x$ and $y$ which is the unique point which is equidistant to $x$ and $y$. By substracting this value, we shift $x$ and $y$ to opposite sides of zero, but we have preserved distances. That is, the point that $x$ gets sent to has the same distance to zero as the point that $y$ gets sent to. Therefore they are negatives of one another (if they are not equal), so the question of which is smaller can now be calculated using the absolute value.
A: $
\renewcommand{\max}{\mathrel{\rm max}}
\renewcommand{\min}{\mathrel{\rm min}}
\newcommand{\abs}[1]{\left| #1 \right|}
$The simplest way I see to prove this is to first use
$$
\abs x = x \max -x
$$
and then use the properties of $\;\max\;$ (and $\;\min\;$).
We start at the most complex side, $\;\frac{x+y-\abs{x-y}}{2}\;$, or rather (to avoid many divisions by 2) we start with its numerator, and calculate as follows:
\begin{align}
& x + y - \left| x - y \right| \\
\equiv & \;\;\;\;\;\text{"express $\;\left| \cdot \right|\;$ in terms of $\;\max\;$"} \\
& x + y - ((x - y) \max (y - x)) \\
\equiv & \;\;\;\;\;\text{"the negation of $\;\max\;$ is $\;\min\;$, with the arguments negated"} \\
& x + y + ((y - x) \min (x - y)) \\
\equiv & \;\;\;\;\;\text{"$\;+\;$ distributes over $\;\min\;$"} \\
& (x + y + y - x) \min (x + y + x - y) \\
\equiv & \;\;\;\;\;\text{"simplify"} \\
& (2 \times y) \min (2 \times x) \\
\equiv & \;\;\;\;\;\text{"$\;z \times\;$ distributes over $\;\min\;$, for $\;z \ge 0\;$"} \\
& 2 \times (y \min x) \\
\end{align}
After division by 2 and using the symmetry of $\;\min\;$, this proves the required $\;\frac{x+y-\abs{x-y}}{2} = x \min y\;$.
