Is my proof of this inequality correct?

[Back2College]

Homework Statement

Prove that |a + b| ≤ |a| + |b|.

Homework Equations

|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|.

 

[jbunniii]

√(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.

 

[Back2College]

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|.