# Does anyone know how to prove this inequality

Does anyone know how to prove the following inequality $$\left|\left|x_{1}\right|-\left|y_{1}\right|\right|+\left|\left|x_{2}\right|-\left|y_{2}\right|\right|\le\left|x_{1}-y_{2}\right|+\left|x_{2}-y_{1}\right|$$for all $x_{1},x_{2},y_{1},y_{2}\in\mathbb{R}$ with $\left|x_{1}\right|\ge\left|x_{2}\right|$ and $\left|y_{1}\right|\ge\left|y_{2}\right|$?

Thanks in advance for any help!

-

The general structure of the statement $-$ in some way $-$ reminds us of the triangle inequality. Therefore it is highly likely that the triangle inequality is the key here.

The triangle inequality states that:
$$\vert a+b\vert\leq\vert a\vert+\vert b\vert$$
If you mess around with this inequaity, the desired result is pretty easy to prove. By some means we are to bring in $\vert x\vert -\vert y\vert$. The first logical step would be to rearrange the inequality as
$$\vert a+b\vert-\vert a\vert\leq\vert b\vert$$. Now, to bring in to bring in $\vert x\vert -\vert y\vert$this we can easily consider the correspondence $a \rightarrow x$ and $b\rightarrow (y-x)$ (a difference). If you look at the inequality, this will neatly reduce it into
$$\vert y\vert-\vert x\vert\le \vert y-x\vert$$ Now if we swap the $x$s and $y$s, we get $$\vert x\vert-\vert y\vert\le \vert x-y\vert$$If this is true, then there is no reason that $$\vert\vert x\vert-\vert y\vert\vert\le \vert x-y\vert$$isn't true.
With this, we can now easily create two separate inequalities as:
$$\vert\vert x_1\vert-\vert y_1\vert\vert\le \vert x_1-y_1\vert$$ and, $$\vert\vert x_2\vert-\vert y_2\vert\vert\le \vert x_2-y_2\vert.$$
If you add these two together, You have $$\left|\left|x_{1}\right|-\left|y_{1}\right|\right|+\left|\left|x_{2}\right|-\left|y_{2}\right|\right|\le\left|x_{1}-y_{1}\right|+\left|x_{2}-y_{2}\right|$$

-
If you add those two together you get $\vert x_1-y_1\vert+\vert x_2-y_2\vert$, not $\left|x_{1}-y_{2}\right|+\left|x_{2}-y_{1}\right|$. –  Byron Schmuland Dec 30 '11 at 21:52
I agree with Byron. –  756 Dec 30 '11 at 21:56
But it is a good introduction to my response since I used it in the very first line. I mean inequality $||x|-|y|| \leq |x-y|$. However I think it can be explained in a more intuitive way. It is the same as $|a-b| \le |a' + b'|$ where $a,b \ge 0$ and $a=+/- \ a,\ b=+- \ b$. On the left, we have a subtraction; on the right we have a subtraction (possibly minus subtraction, which does not matter since we have moduli) or an addition (possibly minus addition), the result of which is obviously bigger (equal to) than the result of subtraction. –  savick01 Dec 30 '11 at 22:10
Don't worry. Making mistakes and correcting them is an important part of doing mathematics. –  Byron Schmuland Dec 30 '11 at 23:00
The most recent edit of this answer sure makes me scratch my head! –  The Chaz 2.0 Apr 8 '12 at 8:06

Let's get rid of some irrelevant things.

If we have $$|x_{1}-y_{1}|+|x_{2}-y_{2}|\le|x_{1}-y_{2}|+|x_{2}-y_{1}|\ (*)$$ for positive numbers satisfying your inequalities, then we are done, since different signs on the right side of the inequality may only enlarge it.

Without loss of generality: $x_1 \leq y_1$. So we have three cases:

• $x_2 \leq x_1 \leq y_2 \leq y_1$,
• $x_2 \leq y_2 \leq x_1 \leq y_1$,
• $y_2 \leq x_2 \leq x_1 \leq y_1$.

All the proofs can be done with pictures (which of course can be rewritten mathematically, but pictures are more clear and faster to prepare (on paper)).

I draw the first situation for you:

You can see that the arrows above the axis (left side of the inequality) are as long as (so no longer than) those below (right side of the inequality), which proves (*) in the first case. In the remaining cases the arrows above would be strictly shorter then the arrows below (but the inequality does not have to be strict since in the picture we draw $x_2$ strictly before $x_1$ etc. and in reality we may have for example $x_1=x_2=y_1=y_2$).

-