# Is my proof of this inequality correct?

## Homework Statement

Prove that |a + b| ≤ |a| + |b|.

|a| = √a2

## The Attempt at a Solution

Since |a| = √a2, then

|a + b| = √(a + b)2 = √(a2 + 2ab + b2) = √a2 + √b2 + √(2ab) = |a| + |b| + √(2ab).

And since the square root of a negative number is not defined, then 2ab must be ≥ 0.

This proves the theorem that |a + b| ≤ |a| + |b|.

## The Attempt at a Solution

jbunniii
Homework Helper
Gold Member
√(a2 + 2ab + b2) = √a2 + √b2 + √(2ab)
Alas, this step is incorrect. To see that this is the case, try ##a=b=1##.

Instead of working with square roots, may I suggest trying to prove this inequality:
$$(|a+b|)^2 \leq (|a|+|b|)^2$$
Then argue that this inequality implies the one you want.

Thanks, jbunniii, for the hint. As you can see, my algebra is a bit rusty. This is what I came up with after your hint.

To prove that |a + b| ≤ |a| + |b|, we will first attempt to prove that (|a + b|)2 ≤ (|a| + |b|)2.

Since (|a + b|)2 is equal to (a + b)2, we have

a2 + b2 + 2ab ≤ (|a| + |b|)2.

This gives us

a2 + b2 + 2ab ≤ a2 + b2 + 2|a||b|

which reduces to

ab ≤ |a||b|

which must be true.

Thus we can take the square root of both sides of the original equation to prove the theorem that |a + b| ≤ |a| + |b|.