# How to justify?

1. Sep 24, 2010

### lovemake1

1. The problem statement, all variables and given/known data

la - bl <= lal + lbl

2. Relevant equations

3. The attempt at a solution

I have prooved this triangle inequality. but please check if there are any obvious errors.

-2ab <= 2labl
a^2 - 2ab + b^2 <= a^2 + 2labl + b^2
( la - bl)^2 <= (lal + lbl)^2

*sqrt both sides
la - bl <= lal + lbl

The problem is now that i have somewhat prooved this inequality, how do i justify it?
In an assignment or on a test I need to write the process of my work.
How would i go aobut justifying this proof?

Heres my shot at it:

ab <= l ab l is true for all real numbers
therefore -2ab <= 2l ab l is the same case.

add a^2, b^2 to both sides and squareroot it.
lal >= a
lbl >= b
therefore lal + lbl >= la - bl

I dont feel very comfortable justifying inequalities, let alone im not very good at it.
are there any ways to justify without feeling too awkard? some keywords i need to be using?

2. Sep 25, 2010

### HallsofIvy

If you have "proved" it then that is a "justification"?

Are you asking how to justify each step?

You first say that $-2ab\le |2ab|$.
Justify that by looking at cases. If $ab\le 0$ then the two sides are equal. If $ab> 0$ then the left side is negative and the right side is positive.

Next you have $a^2- 2ab+ b^2\le a^2+ |2ab|+ b^2$ which is true because you have added the same thing to both sides of the inequality.

Next, $(a- b)^2\le (|a|+ |b|)^2$
Okay, the left side is exactly the same as the left side in the previous inequality but I would recommend adding something to the previous inequality:
$a^2- 2ab+ b^2\le a^2+ |2ab|+ b^2= |a|^2+ 2|a||b|+ |b|^2$
That last equality is true because $x^2= |x|^2$ for any real number and |xy|= |x||y| for any real numbers.

Finally, from $(a- b)^2\le (|a|+ |b|)^2$
you derive $|a- b|\le |a|+ |b|$.

That is true because if x and y are positive numbers and x> y, then x= y+ a for some positive a so $x^2= y^2+ 2ay+ a^2$ so that $x^2$ is equal to $y^2$ plus some positive number.

Yes, that is a perfectly valid proof.