Are My Proofs of Triangle Inequalities Correct?

[tlkieu]

Just wondering if anyone could confirm if I've headed in the right direction with these
(a) Prove the triangular inequality: |x + y| ≤ |x| + |y|.
(b) Use triangular inequality to prove |x − y| ≥ ||x| − |y||.
(c) Show that if |x − a| < c/2 and |y − b| < c/2 then |(x + y) − (a + b)| < c.

So for (a):
∣x+y∣∣^2 = (x+y)^2
= x^2 + 2xy + y^2
= |x|^2 + 2xy + ∣y∣^2
≤ |x|^2 + 2∣xy∣ + ∣y∣^2
= |x|^2 + 2|x|⋅∣y∣ + ∣y∣^2
= (|x|+∣∣y∣∣)^2
Which shows ∣x+y∣ ≤ |x| + ∣y∣

For (b):
I split it into two proofs
In case (1): If |x| ≥ ∣y∣ we have:
∣|x| − ∣y∣∣ = |x|− ∣y∣, and the proof is finished.

For case (2): If ∣y∣ ≥ |x|:
∣y−x∣ ≥ ∣y∣ − |x| = ∣∣y∣ − |x|∣

That is what I have so far and part (c) I'm not too sure how to approach that one

 

[LCKurtz]

Just wondering if anyone could confirm if I've headed in the right direction with these
(a) Prove the triangular inequality: |x + y| ≤ |x| + |y|.
(b) Use triangular inequality to prove |x − y| ≥ ||x| − |y||.
(c) Show that if |x − a| < c/2 and |y − b| < c/2 then |(x + y) − (a + b)| < c.

So for (a):
∣x+y∣∣^2 = (x+y)^2
= x^2 + 2xy + y^2
= |x|^2 + 2xy + ∣y∣^2
≤ |x|^2 + 2∣xy∣ + ∣y∣^2
= |x|^2 + 2|x|⋅∣y∣ + ∣y∣^2
= (|x|+∣∣y∣∣)^2
Which shows ∣x+y∣ ≤ |x| + ∣y∣

For (b):
I split it into two proofs
In case (1): If |x| ≥ ∣y∣ we have:
∣|x| − ∣y∣∣ = |x|− ∣y∣, and the proof is finished.

For case (2): If ∣y∣ ≥ |x|:
∣y−x∣ ≥ ∣y∣ − |x| = ∣∣y∣ − |x|∣

That is what I have so far and part (c) I'm not too sure how to approach that one

They look correct. For the last one take the left side of what you are trying to prove and break it up into two terms similar to what you are given and use the triangle inequality on it.