Inequality involving the min function I'm trying to prove the following inequality: 
$$
\left|y_{1}\land x_{1}-y_{2}\land x_{2}\right|\leq\left|y_{1}-y_{2}\right|+\left|x_{1}-x_{2}\right|,
$$
where $x\land y=\mbox{min}(x,y)$. By considering different cases
it seems to hold but I haven't been able to find a simple way of proving
it. I'd appreciate any suggestions. 
 A: Hint: Consider the four cases:


*

*$y_1<x_1$ and $y_2<x_2$

*$y_1<x_1$ and $y_2\geq x_2$

*$y_1\geq x_1$ and $y_2<x_2$

*$y_1\geq x_1$ and $y_2\geq x_2$
By symmetry, you only need to consider the first two cases.


*

*The first case is simple because the LHS is a term on the RHS.

*For the second case, either $y_1<x_2<y_2$ so that $|y_1-y_2|=|y_1-x_2|+|x_2-y_2|$ or $x_2<y_1<x_1$ so $|x_2-x_1|=|x_2-y_1|+|y_1-x_1|$.  In either case, we get that $|y_1-x_2|$ is less than or equal to a term on the RHS.
A: Here is an exotic, one-line-r based on $$\min\left ( x,y \right )=\frac{1}{2}\left ( x+y - \left | x-y \right | \right )$$
$$\left | \min\left ( x_1,y_1 \right ) - \min\left ( x_2,y_2 \right ) \right |=\left | \frac{1}{2}\left ( x_1+y_1-\left | x_1-y_1 \right | \right ) - \frac{1}{2}\left ( x_2+y_2-\left | x_2-y_2 \right | \right ) \right |=$$
$$=\left | \frac{1}{2}\left ( x_1-x_2 \right ) + \frac{1}{2}\left ( y_1-y_2 \right ) + \frac{1}{2}(\left | x_2-y_2 \right | - \left | x_1-y_1 \right |) \right | \leq$$ 
$$\leq \frac{1}{2}\left | x_1-x_2 \right | + \frac{1}{2}\left | y_1-y_2 \right | + \frac{1}{2}\left | \left | x_2-y_2 \right | - \left | x_1-y_1 \right | \right |\leq $$
$$\leq \frac{1}{2}\left | x_1-x_2 \right | + \frac{1}{2}\left | y_1-y_2 \right | + \frac{1}{2}\left | \left ( x_2-y_2 \right ) - \left ( x_1-y_1 \right ) \right |= $$
$$= \frac{1}{2}\left | x_1-x_2 \right | + \frac{1}{2}\left | y_1-y_2 \right | + \frac{1}{2}\left | \left ( x_2-x_1 \right ) + \left ( y_1-y_2 \right ) \right |\leq$$
$$\leq \frac{1}{2}\left | x_1-x_2 \right | + \frac{1}{2}\left | y_1-y_2 \right | + \frac{1}{2} \left | x_2-x_1 \right | + \frac{1}{2} \left | y_1-y_2 \right | = \left | x_1-x_2 \right | + \left | y_1-y_2 \right | $$
A: (Using the approach from Is $|f(a) - f(b)| \leqslant |g(a) - g(b)| + |h(a) - h(b)|$? when $f = \max\{{g, h}\}$):
If $x_2 \le y_2$ then
$$
 y_{1}\land x_{1}-y_{2}\land x_{2} 
 \le x_1 - x_2 \le \lvert x_1 - x_2\rvert \le \max \{ \lvert x_1 - x_2 \rvert, \lvert y_1 - y_2 \rvert \}
$$
and the same is true if $x_2 \ge y_2$. By symmetry it follows that
$$
 \lvert y_{1}\land x_{1}-y_{2}\land x_{2} \rvert \le
  \max \{ \lvert y_1 - y_2 \rvert, \lvert x_1 - x_2 \rvert \}
$$
which is slightly stronger than the desired
$$
 \lvert y_{1}\land x_{1}-y_{2}\land x_{2} \rvert \le
   \lvert y_1 - y_2 \rvert +  \lvert x_1 - x_2 \rvert \quad .
$$
