# Need help with a proof about absolute value

Prove $$||x| - |y|| \le |x-y|$$

Here's my attempt of the proof:

Since $$x-y = |x-y|$$ or $$x-y = -|x-y|$$, then $$-|x-y| \le x-y \le |x-y|$$.

Also, $$|x| = |-x|$$ and $$|y| =|-y|$$, so we have that $$|x|-|y| \le |x-y|$$.

We know that for any $$\epsilon \gt 0$$, $$|z| \le \epsilon$$ iff $$-\epsilon \le z \le \epsilon$$.

Let $$z = |x|-|y|$$ and take $$\epsilon = |x-y|$$.

This implies $$||x| - |y|| \le |x-y|$$.

I want to know if I missed any key details that I should've written in my proof, and above all, is my proof correct?

• $x-y = |x-y|$ this is not true, better to write $x-y \leq |x-y|$. Similarly $-|x-y|\leq x-y$ – Okkes Dulgerci Mar 11 at 0:20
• By the way -- inequality in the title is known as the reverse triangle inequality. It is discussed on MSE here: math.stackexchange.com/questions/127372/… – Eevee Trainer Mar 11 at 0:23
• Ah thank you, should've checked beforehand. – Ash Mar 11 at 0:24

Ah yes, the reverse triangle inequality, appealing to the intuition insofar as it confirms that the absolute difference in the magnitudes $$\vert x \vert$$, $$\vert y \vert$$ of $$x$$ and $$y$$ is bounded by their difference $$\vert x - y \vert$$ itself: if $$x$$ is close to $$y$$, then the size of $$x$$ is close to the size of $$y$$:

$$\vert \vert x \vert - \vert y \vert \vert \le \vert x - y \vert; \tag 0$$

since our OP Ash's work on proving this has been adequately debriefed in the comments to the question itself, I simply present my own demonstration, merely a re-iteration of a very standard approach:

$$\vert y \vert = \vert x + y - x \vert \le \vert x \vert + \vert y - x \vert = \vert x \vert + \vert x - y \vert; \tag 1$$

so

$$\vert y \vert - \vert x \vert \le \vert x - y \vert; \tag 2$$

we may reverse the roles of $$x$$ and $$y$$ in the above we obtain

$$\vert x \vert - \vert y \vert \le \vert x - y \vert; \tag 3$$

we may multiply (2) by $$-1$$ and find

$$-\vert x - y \vert \le \vert x \vert - \vert y \vert; \tag 4$$

we combine (3) and (4):

$$-\vert x - y \vert \le \vert x \vert - \vert y \vert \le \vert x - y \vert, \tag 5$$

which by definition is equivalent to

$$\vert \vert x \vert - \vert y \vert \vert \le \vert \vert x - y \vert \vert = \vert x - y \vert. \tag 6$$

$$OE\Delta$$.

Your question $$\vert\vert x\vert-\vert y\vert\vert\le\vert x-y\vert$$ Square both sides $$(\vert x\vert-\vert y\vert)^2\le(x-y)^2$$ Distribute $$\vert x\vert^2 + \vert y\vert^2 - 2\vert x\vert\vert y\vert\le x^2+y^2-2xy$$ Simplify $$x^2 + y^2 -2\vert x\vert\vert y\vert\le x^2+y^2-2xy$$ $$x^2+y^2$$ Cancel out $$-2\vert x\vert\vert y\vert\le-2xy$$ Divide by $$-2$$ $$\vert x\vert \vert y\vert\ge xy$$ Which is the same as $$\vert xy\vert\ge xy$$ And to finish this is $$\vert a\vert\ge a$$ Thanks, I had a lot of fun answering this