Reverse triangle inequality proof verification

Question

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!

Answer 1

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 $||x| - |y|| \leq |x - y|$.