The $\min(\cdots ,\cdots)$ function

Question

I have a question (might turn out silly, but I am missing something!) based on the minimum function.

Does $|x-y|+|y-z|\ge |x-z| \implies \min(|x-y|,1)+\min(|y-z|,1)\ge \min (|x-z|,1)\ \ \ ?$

$x,y,z\in \Bbb{R}$

Answer 1

This can be solved by just grinding through the cases:

Suppose $a,b $ are non negative. Then $\min(a+b,1) \le \min(a,1)+\min(b,1)$.

To see this:

If $a+b \le 1$, then both $a,b$ are also, and the formula follows.

If $a+b >1$, then suppose that either of $a,b$ is $\ge 1$, then the formula follows. Otherwise, $a,b$ are both $<1$ in which case $\min(a+b,1) = 1 < a+b = \min(a,1)+\min(b,1)$.

Then we have $\min(|x-z|,1) \le \min(|x-y|+|y-z|,1) \le \min(|x-y|,1)+\min(|y-z|,1)$.

Answer 2

You want to show that $\min(a+b, c) \le \min(a, c)+\min(b, c) $ where $a \ge 0, b \ge 0, c > 0$.

If you use $\min(u, v) =\dfrac{u+v-|u-v|}{2} $, you get (using $c$ instead of $1$) $\min(a+b, c) =\dfrac{a+b+c-|a+b-c|}{2} $ and $\min(a, c)+\min(b, c) =\dfrac{a+c-|a-c|}{2}+\dfrac{b+c-|b-c|}{2} =\dfrac{a+b+2c-|a-c|-|b-c|}{2} $.

So we want $a+b+c-|a+b-c| \le a+b+2c-|a-c|-|b-c| $ or $|a-c|+|b-c| \le c+|a+b-c| $.

It would be nice if this could be proved using the triangle inequality, but I have not been able to do this. So, I will resort to cases, which sort of removes the reason for doing this in the first place. Oh, well.

Assume that $a \ge b$, since this is symmetrical in $a$ and $b$.

If $b \ge c$, this is $a-c+b-c \le c+a+b-c $ or $c \ge 0$, which is true.

If $a \ge c \ge b$, this is $a-c+c-b \le c+a+b-c $ or $b \ge 0$, which is true.

Note that if we allow $b < 0$, the inequality can be false. If $a=2, b=-1$, and $c = 1$, then $\min(a,c)+\min(b, c) =1-1 =0 $ and $\min(a+b, c) =1 $.

Finally, if $c \ge a \ge b$, this is $c-a+c-b \le c+|a+b-c| $ or $c \le a+b+|a+b-c| $.

There are a final two cases depending on how $c$ compares with $a+b$.

If $c \le a+b$, this is $c \le a+b+a+b-c$ or $2c \le 2(a+b)$, which is true.

If $c \ge a+b$, this is $c \le a+b+c-a-b =c $ which is true.

Sort of annoying that I have to go through all these cases.