Proof of ∀(x,y)∈R[|x-y| ≥ |x| - |y|]

Question

Is the following proof correct?

To prove: ∀(x,y)∈R[|x-y| ≥ |x| - |y|]; where R is the set of real numbers.

Proof:

Lemma: ∀(x,y)∈R[|x+y| ≤ |x| + |y|]

Since x and y are arbitrary real numbers we have,

∀(x,y)∈R[|x+(-y)| ≤ |x| + |-y|]

Since |y| = |-y|,

|x - y| ≤ |x| + |y| ⇔ -|x - y| ≥ -|x| - |y|⇒|x - y| ≥ -|x| - |y| ⇔|x - y| + |y| ≥ -|x|

Applying the Lemma we get, |x - y| + |y| ≥ |x - y + y| = |x|

Therefore, |x - y| ≥ |x| - |y|

This concludes the proof.

Answer 1

Or more simply:

Proof. Applying the lemma we get $|x| = |(x-y)+y| \leq |x-y| + |y|$, and then adding $-|y|$ to both sides we get the desired result. $\blacksquare$