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I want to prove that, $$\left| x-y\right|\geq\left| x\right| -\left| y\right|$$ I approached the problem like this af:

Since $-\left|z\right|\leq z\leq\left|z\right|$ is true for any real number $z$, we can write the same for reals $x$ and $y$: $$-\left|x\right|\leq x\leq\left|x\right|$$ $$-\left|y\right|\leq y\leq\left|y\right|$$ We can subtract one expression from the other and get: $$-(\left|x\right|-\left|y\right|)\leq x-y\leq\left|x\right|-\left|y\right|$$

And, since $-a\leq z\leq a$ is equivalent to $\left|z\right|\leq a$ (for any real number $z$), we can write the following: $$\left| x-y\right|\leq\left|x\right|-\left|y\right|$$

This is, in fact, the opposite of what I intended to prove. I know the correct derivation though. Since $-\left|z\right|\leq z\leq\left|z\right|$ is true for any real number $z$, we can write the same for $y$ and $x-y$: $$-\left|y\right|\leq y\leq\left|y\right|$$ $$-\left|x-y\right|\leq x-y\leq\left|x-y\right|$$ By adding the two inequalities, we get: $$-(\left|x-y\right|+\left|y\right|)\leq x\leq\left|x-y\right|+\left|y\right|$$ And, since $-a\leq z\leq a$ is equivalent to $\left|z\right|\leq a$ (for any real number $z$), we can write the following: $$\left| x\right|\leq\left|x-y\right|+\left|y\right|$$ Finally, we can subtract $\left|y\right|$ from both sides and voila: $$\left| x\right|-\left| y\right|\leq\left|x-y\right|$$

This equation is simply failsafe, it holds true for all real values of $x$ and $y$. I'm more curious about $\left| x-y\right|\leq\left|x\right|-\left|y\right|$. It seems algebraically correct but ho, it's experimentally BS. Try placing $x=-6$ and $y=-7$ for instance. The values for $\left| x-y\right|$ and $\left|x\right|-\left|y\right|$ are respectively $1$ and $-1$.

Total BS, no?

So where did I screw up? I must have screwed up algebraically, but where? What's so wrong about $\left| x-y\right|\leq\left|x\right|-\left|y\right|$, algebraically?

HERO
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3 Answers3

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You can't substract inequalities in that way, for example:

$$ +3<4<50$$ $$ -8< 2< 5 $$ If you substract:

$$ 11<2<45 $$ which is wrong.

Maybe you should try this:

$$|x|=|x-y+y|\le |x-y|+|y| \Rightarrow \color{blue}{ |x|-|y|\le |x-y| }$$

going further:

$$|y|=|y-x+x|\le|y-x|+|x|=|x-y|+|x|\Rightarrow \color{blue}{-|x-y|\le |x|-|y|}$$

So:

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

and you can conclude: $$||x|-|y||\le|x-y|$$

L F
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  • I see, I was mistaken from the very beginning. Didn't know I can't subtract inequalities in this way. – HERO Aug 08 '19 at 17:40
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A proof by vectors enter image description here

$$|\vec z|^2 = |\vec x- \vec y|^2 = (\vec x - \vec y) \cdot (\vec x - \vec y) = |\vec x| ^2 + |\vec y|^2 - 2|\vec x \vec y | \le |\vec x|^2 -2|\vec x|\|\vec y| + |\vec y|^2 = (|\vec x| -|\vec y| )^2$$

$$\text{ or } |\vec x - \vec y| \le ||\vec x| - |\vec y||$$

If $\vec{x}$ and $\vec{y}$ are along the same direction, then we may simply write $\vec x \equiv x$ and $\vec y \equiv y$

Thus,

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

Community
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19aksh
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  • Shouldn't you write $\left|\vec{x}-\vec{y}\right|\cdot\left|\vec{x}-\vec{y}\right|$ instead of $(\vec{x}-\vec{y})\cdot(\vec{x}-\vec{y})$? – HERO Aug 03 '19 at 08:55
  • @AmandaMacaurenni The dot product : $$\vec{a}\cdot\vec{a} = |\vec{a}||\vec{a}|\cos0 = |\vec{a}||\vec{a}|$$ . So both are equivalent. – 19aksh Aug 03 '19 at 09:10
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There is a way to prove this inequality using basic algebra (or arithmetic, really). The first case is where $x = y = 0$, so that $|x - y| = |x| - |y| = 0$. The second case is where $|x| < |y|$. Then $|x| - |y| < 0$, and $|x - y|$ equals either $||x| - |y|| < 0$ (when $x$ and $y$ are either both negative or both positive) or $|x| + |y| > 0$ (when $x$ and $y$ have opposite signs). Now, $||x| - |y|| \geq |x| - |y|$ and $|x| + |y| > |x| - |y|$, so we find that $|x - y| \geq |x| - |y|$. For the third case, take the second case, and replace $x$ with $y$ and $y$ with $x$ to obtain the same inequality.

Amy Ngo
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