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Show that $\left| x-y \right| \leq \left| x \right| + \left| y \right|$ for all real numbers $x$ and $y$

By definition, $-|x| \leq x \leq |x|$ and $-|y|\leq y \leq |y|$.

$\Rightarrow -|x|+|y| \leq x-y \leq |x|-|y| \Leftrightarrow -(|x|-|y|) \leq x-y \leq |x|-|y|$

By definition, $|a|<c = -c < a < c$.

$\Rightarrow |x-y| \leq |x|-|y|$

Clearly the signs are different. I only know how to manipulate the triangle inequality which has the same sign on both sides. Here I don't know what to do with different sign on each side.

I suppose I could say $|x|+|y| \geq |x|-|y|$. Would this suffice?

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    $\begingroup$ Well, your version is false. Say $x=0$ and $y=1$. Then you are claiming that $1≤-1$. $\endgroup$
    – lulu
    Commented Jul 8 at 18:07
  • $\begingroup$ Do you have triangle inequality? Or are you proving this from scratch? $\endgroup$
    – Alex Jones
    Commented Jul 8 at 18:08
  • $\begingroup$ Also here: math.stackexchange.com/q/829793/42969 and here: math.stackexchange.com/q/1421512/42969 – all found with Approach0 $\endgroup$
    – Martin R
    Commented Jul 8 at 18:13
  • $\begingroup$ As best I can tell, you have assumed $-|y| \leq y \leq |y|$ implies $|y| \leq -y \leq -|y|$. Do you see why this is wrong? $\endgroup$
    – Brian Moehring
    Commented Jul 8 at 18:19
  • $\begingroup$ A problem was how you subtracted inequalities. You had $-\lvert x\rvert \le x$ and $-\lvert y\rvert \le y$, then incorrectly subtracted them and deduced $-\lvert x\rvert - (-\lvert y\rvert) \le x - y$. Also for the RHS. $\endgroup$
    – peterwhy
    Commented Jul 8 at 18:21

1 Answer 1

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$\left| x-y \right| = |x + (-y)| \leq \left| x \right| + \left| -y \right| = |x| + |y|$, as $|y| = |-y|$ for all $y \in \mathbb R.$

If you want direct then observe that $$|x|^2+|-y|^2+2|x||-y|\geq x^2+(-y)^2-2xy$$ $$\implies(|x|+|y|)^2 \geq (x-y)^2\phantom{a}(\because \forall x\in \mathbb{R};\phantom{;}x^2=|x|^2; |x| = |-x|)$$ $$\implies||x|+|y||\geq |x-y|$$ $$\implies |x-y| \leq |x|+|y|.$$

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  • $\begingroup$ I can't believe it's this simple. $\endgroup$
    – ronald christenkkson
    Commented Jul 8 at 18:15
  • $\begingroup$ Yes, it is just an observation or a quick proposition after the main result. $\endgroup$
    – Afntu
    Commented Jul 8 at 18:18
  • $\begingroup$ Dear @Afntu sir, is there any method to communicate to you via email or otherwise ? $\endgroup$
    – Sahaj
    Commented Jul 8 at 18:46
  • $\begingroup$ @Sahaj see my MSE profile. $\endgroup$
    – Afntu
    Commented Jul 9 at 4:42

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