Intuition behind $ ||x|-|y|| \leq |x-y| $

Question

In my math analysis course we proved that $ ||x|-|y|| \leq |x-y| $. However, unlike the triangle equality that I can visualise with the example of a triangle, I cannot visualize this one nor understand why is it true.. So what would be an intuitive way to think about this inequality?

Answer 1

I would think about it as the distance between two points. If you consider $-5$ and $2$ on the real number line, the distance between those two points is $7$, but if we first take the absolute value of both numbers, we find that the distance is only $3$.

Answer 2

Essentially this inequality expresses the observation that the difference between the absolute values of two numbers does not exceed the actual distance between the two numbers.

If two numbers lie on the same side of $0$ on the number line, then the distance between them is the same as the distance between their absolute values. And if they lie on opposite sides of $0$, their absolute values will be closer together than the two numbers themselves.

Answer 3

This is the triangle inequality in disguise. By symmetry, the inequality is equivalent to $|x|-|y|\leq|x-y|$. We can then add $|y|$ to both sides to get $|x|\leq|x-y|+|y|$. This is the same as $|a|+|b|\geq|a+b|$ by letting $a=x-y$ and $b=y$. So you can visualize it just like the triangle inequality by making those substitutions.

Answer 4

All the following stuff actually holds for vectors $x, y \in V$, a normed space with norm $\Vert \cdot \Vert$; then the inequality of interest actually looks like

$\vert \Vert x \Vert - \Vert y \Vert \vert \le \Vert x - y \Vert. \tag 1$

Consider what $(1),$ which is a well-known consequence of the triangle inequality, tells us: it tells us that the difference in length between the two vectors $x$ and $y$ is bounded by the magnitude of the difference between the vectors themselves; so as two vectors "get closer together", their lengths do as well, and of course this makes a lot of sense, intuitively speaking.

Suppose

$\Vert x \Vert \ne 0 \ne \Vert y \Vert; \tag 2$

then there is some $0 < \alpha \in \Bbb R$ with

$\Vert y \Vert = \alpha \Vert x \Vert; \tag 3$

thus

$\vert \Vert y \Vert - \Vert x \Vert \vert = \vert \vert \alpha \vert \Vert x \Vert - \Vert x \Vert \vert = \vert \alpha - 1 \vert \Vert x \Vert; \tag 4$

therefore,

$\Vert y - x \Vert \ge \vert \alpha - 1 \vert \Vert x \Vert, \tag 5$

so as $\alpha$ grows, the distance between the points $x$ and $y$ grows as well; the intuition here is that the two points $x, y \in V$ cannot remain close when one gets much farther from the origin than the other.

That's the best I can do, "intuitively" speaking.

Nota Bene: Of course the inequality (1) follows directly from the triangle inequality

$\Vert x + y \Vert \le \Vert x \Vert + \Vert y \Vert, \tag 6$

which holds in any normed space such as $V$, since it implies

$\Vert x \Vert = \Vert y + (x - y) \Vert \le \Vert y \Vert + \Vert x - y \Vert, \tag7$

whence

$\Vert x \Vert - \Vert y \Vert \le \Vert x - y \Vert; \tag 8$

interchanging the roles of $x$ and $y$ yields

$\Vert y \Vert - \Vert x \Vert \le \Vert x - y \Vert, \tag 9$

or

$-\Vert x - y \Vert \le \Vert x \Vert - \Vert y \Vert; \tag{10}$

by the definition of $\vert \cdot \vert$, (8) and (10) together imply

$\vert \Vert x \Vert - \Vert y \Vert \vert \le \Vert x - y \Vert,\tag{11}$

as desired. End of Note.

Answer 5

You will find below a vizualisation of the slightly more general property :

$$\underbrace{||x-a|-|x-b||}_{g(x)} \leq |a-b|$$

Explanation : function $g$ represents the "gap" between the two "curves" (= broken lines) which are graphical representations of $y=|x-a|$ and $y=|x-b|$.

Two cases (the first one being dominant) for this gap, represented by double arrows :

• either $x \in (a,b)$ : in this case, this gap is less that the quantity $|a-b|$, being even zero in the exceptional case where $x=\frac{a+b}{2},$

• or $x \leq a$ or $b \leq x$ : in this case, this gap is constant and equal to $|a-b|$ ; this case represents the vast majority of cases.

Otherwise said, the most usual situation in (1) is a case of equality !

Fig. 1 : The gap between $|x-a|$ and $|x-b|$ (represented by vertical double arrows) is never more than $|a-b|$, a quantity that is "readable" on the $x$ and on the $y$ axis (sides of the dotted square).

Answer 6

The standard triangle inequality says the longest side of a triangle can't be longer than the sum of the lengths of the other two sides. This alternate form says the shortest side must be longer than the difference between the other two. Imagine folding up the two longest sides like a jackknife; if the shortest side is less than or equal to the difference between them, the whole triangle collapses ino a line segment.