A better way to prove $|x - w| + |y - z| \geq |x - z| + |y - w|$ if $x\leq y$ and $z\leq w$

Question

Let $x\leq y$ and $z\leq w$. Prove that:

$|x - w| + |y - z| \geq |x - z| + |y - w|$

This can be proved by taking cases for $y$ and using the triangle inequality:

for $y \geq w$ :

$|x - w| + |y - z| \geq |x - z| + |y - w|$

$|x - w| + (y - z) \geq |x - z| + (y - w)$

$(w - z) + |y - z| \geq |y - w|$

$|w - z| + |y - z| \geq |y - w|$ (true from the triangle inequality)

Then work similarly for the cases $z \leq y < w$ and $y < z$

I'm wondering if there is a way to prove that which doesn't involve as many cases and operations.

Answer 1

Without loss of generality and assuming $w\geq z$ and $y\geq x$, we may assume that one of $y$ and $w$ is the leftist (most left) point on the real line. Say $w$ is the leftist point. Then,

$|x - w| + |y - z| \geq |x - z| + |y - w|\iff x - w + |y - z| \geq |x - z| +y - w\iff$

$x-y + |y - z| \geq |x-z|\iff |x-y| + |y - z| \geq |x-z|$

which is true by triangular inequality.

Answer 2

Slightly more general one can show that if $x\leq y$ and $z\leq w$ then $$ f(x - w) + f(y - z) \geq f(x - z) + f(y - w) $$ for any convex function $f: \Bbb R \to \Bbb R$. Your inequality is a special case because $f(t) = |t|$ is convex.

For a proof let $$ a = x-w, b= x-z, c = y-w, d= y-z \, . $$ Then $$ a \le b \le d \, , \\ a \le c \le d \, $$ and $$ a+d = b+c \, . $$ If $a=d$ then $a=b=c=d$ and we are done. Otherwise $$ f(b) \le \frac{d-b}{d-a}f(a) + \frac{b-a}{d-a}f(d) \\ f(c) \le \frac{d-c}{d-a}f(a) + \frac{c-a}{d-a}f(d) \\ $$ and adding these inequalities gives $$ f(b) + f(c) \le f(a) + f(d) \, . $$

This is also a special case of Karamata's inequality with $$ \begin{align} (x_1, x_2) &= (d, a) \, ,\\ (y_1, y_2) &= (\max(b,c), \min(b, c) \, . \end{align} $$

Answer 3

We assume here that $x\le y$ and $z\le w$.

The function $f(t) = |t-w| - |t-z|$ is continuous and it is differentiable except at points $w$ and $z$ with \begin{equation} f'(t) = \text{sgn}(t-w)-\text{sgn}(t-z) \le 0 \end{equation} because $\text{sgn}()$ is non-decreasing and $t-w\le t-z$.

It follows that $f$ is non-increasing in $\mathbb{R}$, hence $f(x)\ge f(y)$, i.e. \begin{equation} |x-w|-|x-z|\ge |y-w|-|y-z|\tag*{$\blacksquare$} \end{equation}