# Very Simple Inequalities Proof

1. Jun 24, 2012

### clarence829

1. The problem statement, all variables and given/known data

Prove the following inequalities for all numbers x, y.

|x+y| ≥ |x|-|y|

[Hint: Write , and apply , together with the fact that

2. Relevant equations

These were given as hints in my textbook:

x=x+y-y
|a+b| ≤ |a| + |b|
|-y|=|y|

3. The attempt at a solution

I realize that this is very elementary but this is my first day teach myself calculus and that I am inexperienced using proofs in math. Any and all help is greatly appreciated.

1) x=x+y-y
2) |x+y|≤|x|+|y|
3) |-y|=|y|
4) x+y=x+y
5) √(x+y)^2=√(x+y)^2
6) |x+y|=|x+y|
7) |x+y|≤|x|+|y|
8) |x+y|≤|x|+|-y|

2. Jun 24, 2012

### klondike

Can you see |x+y|+|-y|>= ? by the relevant facts you were given?

3. Jun 24, 2012

### clarence829

If I could get to |x+y|+|-y|≥ |x| then I'd just have to pull |-y| to the right side of the inequality and apply |-y|=|y| to |-y| and I'd have my proof. The problem is that I'm unfamiliar with the various algebraic rules that apply to absolute value, particularly in inequalities.

What are the rules and steps that will get me from x=x+y-y to ≤|x+y|+|-y≥|x|?

4. Jun 24, 2012

### SammyS

Staff Emeritus
To get to |x+y|+|-y|≥ |x|, use |a+b| ≤ |a| + |b| which is equivalent to |a| + |b| ≥ |a+b|.

x+y takes the role of a.

-y takes the role of b.

5. Jun 24, 2012

### clarence829

@SammyS

I see how you're getting your solution and I appreciate the help.

After the question in my textbook (Lang) it says "Hint: Write x=x+y-y, and apply |a+b| ≤ |a|+|b|, together with the fact that |-y|=|y|.

How would one use x=x+y-y in solving this proof?

6. Jun 24, 2012

### SammyS

Staff Emeritus
x = x+y-y

|x| = |(x+y)+(-y)| ≤ |(x+y)| + |(-y)| ...

7. Jun 24, 2012

### clarence829

Everything just finally clicked and it all makes sense now. Thanks again for the help.

8. Jun 24, 2012

### SammyS

Staff Emeritus
You're welcome !