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?

  • $\begingroup$ $x-y = |x-y|$ this is not true, better to write $x-y \leq |x-y|$. Similarly $-|x-y|\leq x-y $ $\endgroup$ – Okkes Dulgerci Mar 11 at 0:20
  • 3
    $\begingroup$ 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/… $\endgroup$ – Eevee Trainer Mar 11 at 0:23
  • $\begingroup$ Ah thank you, should've checked beforehand. $\endgroup$ – 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:

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


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

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

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



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


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