# Simply inequality proof

1. Aug 26, 2013

### phospho

prove $|a+b| \leq |a| + |b|$

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

$(|a+b|)^2 = (a+b)^2 = a^2 + 2ab + b^2$
$\leq a^2 + 2|a||b| + b^2$
$= |a|^2 + 2|a||b| + |b|^2$
$= (|a|+|b|)^2$

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

I don't get there the $\leq$ comes from... nor the conclusion

2. Aug 26, 2013

### Curious3141

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?

3. Aug 26, 2013

### phospho

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

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

$\leq a^2 + 2|a||b| + b^2$

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

thanks

4. Aug 26, 2013

### Curious3141

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.

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.

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.

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

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.

5. Aug 26, 2013

The first $\le$ occurs because, for any numbers $w$, $w \le |w|$

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

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

6. Aug 26, 2013

### phospho

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