# Is this an acceptable proof?

Agent M27

## Homework Statement

Let a & b be real numbers.

Prove that:
|a+b|<=|a|+|b|

## Homework Equations

|x|=$$\sqrt{x^{2}}$$

## The Attempt at a Solution

|a+b|

=$$\sqrt{(a+b)^{2}}$$

=$$\sqrt{(a^{2}+2ab+b^{2})}$$ <= $$\sqrt{a^{2}} + [tex]\sqrt{b^{2}}$$

|a|=$$\sqrt{a^{2}}$$

|b|=$$\sqrt{b^{2}}$$

I feel like this is lacking in foundation, but I lack in the foundation of proofs involving absolute value. Thanks in advance for the assistance.

Joe

Sorry for the ugly formatting, tex is cumbersome sometimes.

Last edited:

Staff Emeritus
Homework Helper
This is ok. But I would like some more details in the following inequality:

$$\sqrt{a^2+2ab+b^2}\leq\sqrt{a^2}+\sqrt{b^2}$$

It's not immediately obvious why this should be true...

Homework Helper

## Homework Statement

Let a & b be real numbers.

Prove that:
|a+b|<=|a|+|b|

## Homework Equations

|x|=$$\sqrt{x^{2}}$$

## The Attempt at a Solution

|a+b|

=$$\sqrt{(a+b)^{2}}$$

=$$\sqrt{(a^{2}+2ab+b^{2})}$$ <= $$\sqrt{a^{2}} + \sqrt{b^{2}}$$
This is the only place I see a difficulty. How do you prove this? I suspect it is no simpler to prove this than it is to prove $|a+ b|\le |a|+ |b|$ by "considering cases": a positive or negative, b positive or negative.

|a|=$$\sqrt{a^{2}}$$

|b|=$$\sqrt{b^{2}}$$

I feel like this is lacking in foundation, but I lack in the foundation of proofs involving absolute value. Thanks in advance for the assistance.

Joe

Sorry for the ugly formatting, tex is cumbersome sometimes.

╔(σ_σ)╝
This is ok. But I would like some more details in the following inequality:

$$\sqrt{a^2+2ab+b^2}\leq\sqrt{a^2}+\sqrt{b^2}$$

It's not immediately obvious why this should be true...

I completely agree. In fact, I have never seen this before.

How can you prove this ?

## Homework Statement

Let a & b be real numbers.

Prove that:
|a+b|<=|a|+|b|

## Homework Equations

|x|=$$\sqrt{x^{2}}$$

## The Attempt at a Solution

|a+b|

=$$\sqrt{(a+b)^{2}}$$

=$$\sqrt{(a^{2}+2ab+b^{2})}$$ <= $$\sqrt{a^{2}} + [tex]\sqrt{b^{2}}$$

|a|=$$\sqrt{a^{2}}$$

|b|=$$\sqrt{b^{2}}$$

I feel like this is lacking in foundation, but I lack in the foundation of proofs involving absolute value. Thanks in advance for the assistance.

Joe

Sorry for the ugly formatting, tex is cumbersome sometimes.

With vectors the best way is to use property of the inner product.

For this however I would simply consider the cases of signs. Because these are numbers in one dimension, you only have to check a few cases.

If sgn(a) = sign(b) then |a + b| = |a| + |b| if a,b >= 0. |a + b| = |-c - d| = |-(c+d)|
= |-c| + |-d|

Now let sgn(a) = 1 - sgn(b)

Then |a + b| < |a| + |b| can be proved simply from the property

|a + b|^2 = a^2 + b^2 + 2ab

But if sgn(a) = 1 - sgn(b) then 2ab < 0

So |a+b|^2 < |a|^2 + |b|^2 from above property