How is the inequality |a+b| \leq |a| + |b| proven using different methods?

[phospho]

prove [itex] |a+b| \leq |a| + |b| [/itex]

i've proved it considering all the 4 cases for a and b but the book went about it a different way:

[itex] (|a+b|)^2 = (a+b)^2 = a^2 + 2ab + b^2 [/itex]
[itex] \leq a^2 + 2|a||b| + b^2 [/itex]
[itex] = |a|^2 + 2|a||b| + |b|^2 [/itex]
[itex] = (|a|+|b|)^2 [/itex]

it then goes on the conclude that [itex] |a+b| \leq |a| + |b| [/itex] because [itex] x^2 \leq y^2 [/itex] implies [itex] x < y [/itex]

I don't get there the [itex] \leq [/itex] comes from... nor the conclusion

 

[Curious3141]

it then goes on the conclude that [itex] |a+b| \leq |a| + |b| [/itex] because [itex] x^2 \leq y^2 [/itex] implies [itex] x < y [/itex]

Should be ##x^2 \leq y^2 \implies x \leq y##.

And it only holds for non-negative ##x## and ##y##. Not a problem here, because in the original proof, you're dealing with absolute quantities which are, by definition, non-negative.

Prove it by bringing the ##y^2## to the LHS, then factorise. What can you conclude?

 

[phospho]

Should be ##x^2 \leq y^2 \implies x \leq y##.

And it only holds for non-negative ##x## and ##y##. Not a problem here, because in the original proof, you're dealing with absolute quantities which are, by definition, non-negative.

Prove it by bringing the ##y^2## to the LHS, then factorise. What can you conclude?

## x^2 - y^2 \leq 0 ##
## (x+y)(x-y) \leq 0 ##
## -y \leq x \leq y ##

I don't understand what that shows, also I don't this step in their proof:

[itex] \leq a^2 + 2|a||b| + b^2 [/itex]

where did the ## \leq ## sign come from? Also, their proof is for [itex] |a+b| = |a| + |b| [/itex], how does their conclusion make it ## |a+b| \leq |a| + |b| ## ?

thanks

 

[Curious3141]

## x^2 - y^2 \leq 0 ##
## (x+y)(x-y) \leq 0 ##

From this step, observe that ##(x+y)## is non-negative (because ##x## and ##y## are both non-negative). Hence you can cancel ##(x+y)## from the LHS and you're left with ##x \leq y## as required.

I don't understand what that shows

That allows you to go from ##(|a + b|)^2 \leq (|a| + |b|)^2## to ##|a + b| \leq |a| + |b|##, which is the very thing they are trying to prove.

, also I don't this step in their proof:

[itex] \leq a^2 + 2|a||b| + b^2 [/itex]

Basically, that amounts to saying ##ab \leq |a||b|##. The equality holds as long as both a and b have the same sign (either both positive or both negative) or when at least one of them is zero. But if a and b have opposite sign (i.e. one of them is positive, the other is negative), then ##ab < |a||b|## because the LHS will become negative while the RHS is always positive.

So ##ab \leq |a||b|## covers all the possibilities. Multiply that by 2 to get ##2ab \leq 2|a||b|##. Add ##a^2 + b^2## to it to get exactly what they wrote down.

Also, their proof is for [itex] |a+b| = |a| + |b| [/itex]

No it isn't and that statement is not even true in general. Why do you think that?

how does their conclusion make it ## |a+b| \leq |a| + |b| ## ?

Work through the algebra again. I think you're getting confused because the signs alternate between equal and less-than-equal, then back to equal in a long "chain". I think you should write all the steps down systematically in terms of a single LHS and RHS with only a single sign between them at each step, and see if you can understand it better.

 

[statdad]

The first [itex] \le [/itex] occurs because, for any numbers [itex] w [/itex], [itex] w \le |w| [/itex]

Next, you have reached the point where you have
[tex]
|a+b|^2 \le \left( |a| + |b| \right)^2
[/tex]

Everything involved is non-negative because of the absolute values: what happens when you apply square roots?

 

[phospho]

thank you both... I wrote it down step by step and got it.