# Reverse triangle inequality proof verification

Whilst trying to come up with a proof for the reverse triangle inequality, I came up with this. I know that I overcomplicated things, but still wanted to know whether what I did was correct.

What I need to prove is that $$\forall x, y \in \mathbb{R}: |x-y|\geq ||x|-|y||$$. I tackled this problem with a proof by contradiction. So, assume that $$\forall x, y\in \mathbb{R}: |x-y| < ||x|-|y||$$.

Then I squared both sides in order to get rid of the absolute values. Because both the left and right hand side of the inequality are positive, the inequality stays the same.

$$(x-y)^2 < (|x|-|y|)^2$$

$$x^2 - 2xy + y^2 < x^2 - 2|xy| + y^2$$

Simplifying this gives

$$-2xy < -2|xy|$$

Division by a negative number, so inequality changes

$$xy > |xy|$$

Which can never be true because $$\forall a \in \mathbb{R}: |a| \geq a$$, thus we have a contradiction and so $$\forall x, y \in \mathbb{R}: |x-y|\geq ||x|-|y||$$ must be true.

I'm fairly new to writing my own proofs, so any help would be much appreciated!

• Your proof looks good to me, and I do not think it's overly complicated. I would even say it's more elegant than the brute force one from the Wikipedia page :) – janosch Aug 1 '19 at 13:25

I think your proof is correct. There is an easier proof though. Notice that $$|x| = |x - y + y| \leq |x - y| + |y|$$ by the triangle equality. This expression rearranges to $$|x| - |y| \leq |x - y|$$. Similarly, $$|y| = |y - x + x| \leq |y - x| + |x| = |x - y| + |x|$$, which rearranges to $$|y| - |x| \leq |x - y|$$. Multiplying by -1 yields $$|x| - |y| \geq -|x - y|$$. Combining this inequality with the previous one yields $$-|x - y| \leq |x| - |y| \leq |x - y|$$, or $$||x| - |y|| \leq |x - y|$$.