Is $|x_1-y_2|^p+|x_2-y_1|^p \ge |x_1-y_1|^p+|x_2-y_2|^p$ for any $0 \le x_1 \le x_2$ and $0 \le y_1 \le y_2$ any positive integer $p$? I think it holds for $p=1$, since the only possible cases are
$x_1\le x_2 \le y_1 \le y_2$
$x_1\le y_1 \le y_2 \le x_2$
$x_1\le y_1 \le x_2 \le y_2$
and their symmetric ones (exchange $x$ and $y$), and we can draw on real lines to show the inequality holds for $p=1$ for all such cases.
My question is that is this true for all positive integers $p$ and how do I algebraically prove this? In addition, if this does not hold for any non-negative real number, does it hold for non-negative integers? Thank you!
My attempt (help check if my attempt is correct, thanks!):
For the last case $x_1\le y_1 \le x_2 \le y_2$, let >$|x_1-y_1|=a,|y_1-x_2|=b,|x_2-y_2|=c$, then
$|x_1-y_1|^p+|x_2-y_2|^p=a^p+c^p\le|x_1-y_2|^p+|x_2-y_1|^p=(a+b+c)^p+b^p$, where $a,b,c\ge 0$.
For the second case $x_1\le y_1 \le y_2 \le x_2$, let >$|x_1-y_1|=a,|y_1-y_2|=b,|y_2-x_2|=c$, then
$|x_1-y_1|^p+|x_2-y_2|^p=a^p+c^p\le|x_1-y_2|^p+|x_2-y_1|^p=(a+b)^p+(b+c)^p$, where $a,b,c\ge 0$.
The remaining one is the first case, which is equivalent to ask if $(a+b)^p+(b+c)^p \le (a+b+c)^p + b^p$ for any $a,b,c \ge 0$. This is true since the expansion of $(a+b+c)^p$ contains $(a+b)^p$ and $(b+c)^p$ when $p$ is a positive integer.