Absolute inequality derivation

Question

I have been trying to prove an inequality that I am not even sure if it is even true or not. However I am experiencing great difficulties with this proof. I have an intuition that it is true and have been trying to derive it but with no success. Here is the following set up:

$0 < x \leq y $

$0 < a \leq b $

From this I am trying to derive the following:

$|y-b|+|x-a| \leq |y-a| + |x-b|$

I have tried 'reverse engineering' and trying to come back to a statement that be easily derived from the initial set up but without success. After this I tried playing around with triangle inequalities but ended doing recursive steps in my solution attempts.

The inequality isn't random but some sort of "intuition" coming from that if we add the difference of the smaller numbers (x,a) to the difference of the bigger numbers (y,b) then this will be smaller than if we just pair them up the other way (y,a) and (x,b). Of course this intuition could be wrong.

Does anyone know if this is true and how would I work to prove it?

Answer 1

To prove/disprove $|y-b|+|x-a| \leq |y-a| + |x-b|$, we shall consider 6 cases.

These are the cases:

$$x\le y\le a\le b$$$$x\le a\le y\le b$$$$a\le x\le y\le b$$$$a\le x\le b\le y$$$$x\le a\le b\le y$$$$a\le b \le x \le y$$

For Case 1,

$|y-b|+|x-a|=b-y+a-x=(b-x)+(a-y)\le|y-a|+|x-b|$.

For Case 2,

$|y-b|+|x-a|=b-y+a-x=(b-x)+(a-y)=(b-x)-(y-a)\\=(b-x)-|y-a|\le|x-b|+|y-a|$.

May try case 3-6 on your own.

Answer 2

The equation $$ D((x, y), (x_0, y_0)) = \lvert x - x_0 \rvert + \lvert y - y_0 \rvert = C $$ defines a diamond $D$ with diagonal lenght $2C$ centered around $(x_0, y_0)$.

The image gives the diamonds centered around $(1,2)$ and $(2,1)$ for $C=1$ and $C = 2$. The centers are mirrored along the identity function $y = x$.

• The condition $0 < x \le y$ restricts $(x,y)$ to all points on or above id, for positive $x$.

• The condition $0 < a \le b$ restricts the center $(a,b)$ of the first diamond to that region as well.

Your inequality

$|y-b|+|x-a| \leq |y-a| + |x-b|$

can be stated as $$ D((x,y),(a, b)) \le D((x,y),(b, a)) \quad (*) $$

If $D$ is interpreted as a distance, it states that someone at $(x,y)$ is either closer to the center $(a, b)$ than to the mirrored center $(b,a)$ or at the same distance to both.

This holds on the considered area.

Answer 3

Hint:

$t\mapsto |t|$ is convex, so Karamata's inequality allows you to conclude from $(y-a, x-b)\succ (y-b, x-a)$.