# Triangle Inequality Proof help

• lovemake1
Yes, it is reasonable and valid to make those three cases. It shows that you have thought about the possible scenarios and have accounted for all of them in your proof. Just make sure to explain why you are making those cases and how they contribute to the overall proof.
lovemake1

## Homework Statement

1] l x + y l < or equal to l x l + l y l

x^2 + 2xy + y^2

## The Attempt at a Solution

Left side. i Squared left side to begin with, and i got x^2 + 2xy + y^2 and also did the same for the right side, but it would have absolute sign between each variable.

x^2 + 2xy + y^2 < or equal to lxl^2 + 2lxllyl + lyl^2

so its just (x + y)^2 < or equal to ( lxl+lyl )^2
not sure what I am supppose to do next..

it just seems like I am doing somethign and undoing something without any purpose.
can anyone clear this from my head?

The point is that 2xy<=2|x||y| and x^2=|x|^2 and y^2=|y|^2.

lA - Bl <\ lAl + lBl

We would get the same as above but except it would be Sqrt[(a - b)^2] <\ lAl + lBl

or is it Sqrt[(a - b)^2] = lAl + lBl

lovemake1 said:

lA - Bl <\ lAl + lBl

We would get the same as above but except it would be Sqrt[(a - b)^2] <\ lAl + lBl

or is it Sqrt[(a - b)^2] = lAl + lBl

If you are supposed to prove |A-B|<=|A|+|B|, the final answer would be an argument proving it. Can you give a proof?

lA-Bl^2 <= ( lAl +lBl ) ^2

A^2 - 2AB + B^2 <= lAl^2 + 2lAllBl + lBl^2

-2AB <= 2lAllBl

would this be correct?
im not sure if i can subtract absolute value to a non absolute value...
still confused.

lovemake1 said:
lA-Bl^2 <= ( lAl +lBl ) ^2

A^2 - 2AB + B^2 <= lAl^2 + 2lAllBl + lBl^2

-2AB <= 2lAllBl

would this be correct?
im not sure if i can subtract absolute value to a non absolute value...
still confused.

-2AB is either equal to 2|A||B| or -2|A||B| depending on the signs of A and B, right? In either case, -2AB is less than or equal to 2|A||B|. Is that the part you are asking about?

yea doesn't that prove |A-B|<=|A|+|B|

square both sides.
lA-Bl^2 <= ( lAl +lBl ) ^2

A^2 - 2AB + B^2 <= lAl^2 + 2lAllBl + lBl^2

you subtract A^2 and B^2 from both sides.
your left with -2AB <= 2lAllBl which is true.

or is this not the final stage?

lovemake1 said:
yea doesn't that prove |A-B|<=|A|+|B|

square both sides.
lA-Bl^2 <= ( lAl +lBl ) ^2

A^2 - 2AB + B^2 <= lAl^2 + 2lAllBl + lBl^2

you subtract A^2 and B^2 from both sides.
your left with -2AB <= 2lAllBl which is true.

or is this not the final stage?

Sure. Except to really present it as a proof you probably want to start with -2AB<=2|A||B|. Now add A^2 and B^2 to both sides, factor, note both sides are nonnegative and take the square root. In other words do the derivation backwards. But you certainly have the right idea.

what do you mean by adding a^2 and b^2 to both sides, you mean remove the absolute sign from A and B from the right side?

so my solution above was not good enough as a proof right ?
I need to make sure that my answer is at a reasonable level to proove the inequality.

so now i add A^2 and B^2.
A^2 - 2AB + B^2 <= A^2 + 2lAllBl + B^2

and take the square root ? how?

lovemake1 said:
what do you mean by adding a^2 and b^2 to both sides, you mean remove the absolute sign from A and B from the right side?

so my solution above was not good enough as a proof right ?
I need to make sure that my answer is at a reasonable level to proove the inequality.

so now i add A^2 and B^2.
A^2 - 2AB + B^2 <= A^2 + 2lAllBl + B^2

and take the square root ? how?

A^2=|A|^2. Ditto for B. Now FACTOR both side into squares. Then take the square root.

A^2 - 2AB + B^2 <= A^2 + 2lAllBl + B^2
(A-B)^2 <= (lAl+lBl)^2
(A-B) <= lAl + lBl

uh, now we're back to the original inequality, is this what we're suppose to do ?
do somethign and undo it again ? I am new to prooving inequalities ,
is this correct?

lovemake1 said:
A^2 - 2AB + B^2 <= A^2 + 2lAllBl + B^2
(A-B)^2 <= (lAl+lBl)^2
(A-B) <= lAl + lBl

uh, now we're back to the original inequality, is this what we're suppose to do ?
do somethign and undo it again ? I am new to prooving inequalities ,
is this correct?

It's not a question of doing and undoing something. You 'discovered' the proof by squaring both sides and reducing it to -2AB<=2|A||B|. Which you know is true. So now to prove it you start with what you know is true. Your first line should be -2AB<=2|A||B|. 'Proofs' usually start with what you know, then end up with what you want to prove.

-2AB <= 2lAllBl
A^2 - 2AB + B^2 <= A^2 + 2lAllBl + B^2
(A - B)^2 <= (lAl + lBl)^2
(A - B) <= lAl + lBl

So it would look something like this ?

lovemake1 said:
-2AB <= 2lAllBl
A^2 - 2AB + B^2 <= A^2 + 2lAllBl + B^2
(A - B)^2 <= (lAl + lBl)^2
(A - B) <= lAl + lBl

So it would look something like this ?

It would look exactly like that. If you want to be fancy you could add a comment to each line justifying the step you took. But I think that's good enough.

awesome, thanks for the awesome help !
i just didnt grasp the concept of prooving inequality at first.

lets say this question was in an our midterm exam.
how would i go about justifying ?

Heres my shot at it.

Since we know that 2lAlBl >= 0 for any real number. Therefore, it is greater than
-2AB which yields a negative value for the following 3 cases.

A < 0 , B > 0
A > 0, B < 0
A >0, B >0

By adding A^2 and B^2 to both sides, we can factor and squareroot the equation.
which gives us (A - B) <= lAl + lBl

(One question, is it resonable to just start with -2AB <= 2lAllBl ? )
curious..

lovemake1 said:
lets say this question was in an our midterm exam.
how would i go about justifying ?

Heres my shot at it.

Since we know that 2lAlBl >= 0 for any real number. Therefore, it is greater than
-2AB which yields a negative value for the following 3 cases.

A < 0 , B > 0
A > 0, B < 0
A >0, B >0

By adding A^2 and B^2 to both sides, we can factor and squareroot the equation.
which gives us (A - B) <= lAl + lBl

(One question, is it resonable to just start with -2AB <= 2lAllBl ? )
curious..

-2AB<=2|A||B| in ALL cases. Not just the three you listed. It's 'greater than or equal to' not just 'greater than'.

ok so what is it that I am justifying at the end or in the middle.

either 2lalbl >= -2ab

or that a - b <= lal + lbl

we start with 2lalbl >= -2ab and end with a - b <= lal + lbl, which is what its asking us to proove.

lovemake1 said:
ok so what is it that I am justifying at the end or in the middle.

either 2lalbl >= -2ab

or that a - b <= lal + lbl

we start with 2lalbl >= -2ab and end with a - b <= lal + lbl, which is what its asking us to proove.

That might depend on what you are trying to prove. I'm getting all confused. You started out trying to prove |a+b|<=|a|+|b|, right? Did the problem change? Why don't you state what you are trying to prove and the steps you might use to get there and I'll see if I can help.

Last edited:
yea i started out trying to prove la+bl <= lal + lbl

this is exactly how i solved it.

"
(a-b)^2 <= (lal + lbl)^2
a^2 - 2ab + b^2 <= lal^2 + 2lalbl + lbl^2
a^2 - 2ab + b^2 <= a^2 + 2labl + b^2
Examine the value of -2ab and 2labl. 2labl >= -2ab is true for all real
values since labl >= ab

Take the square root of a^2 - 2ab + b^2 <= a^2 + 2labl + b^2
we have l a - b l <= lal + lbl
"

That is what I had in mine. I wrote it as if this question was on the exam or an assignment. Please i need to know where i went wrong, starting from the beggining of quotation, please let me know whre to fix my errors.

lovemake1 said:
yea i started out trying to prove la+bl <= lal + lbl

this is exactly how i solved it.

"
(a-b)^2 <= (lal + lbl)^2
a^2 - 2ab + b^2 <= lal^2 + 2lalbl + lbl^2
a^2 - 2ab + b^2 <= a^2 + 2labl + b^2
Examine the value of -2ab and 2labl. 2labl >= -2ab is true for all real
values since labl >= ab

Take the square root of a^2 - 2ab + b^2 <= a^2 + 2labl + b^2
we have l a - b l <= lal + lbl
"

That is what I had in mine. I wrote it as if this question was on the exam or an assignment. Please i need to know where i went wrong, starting from the beggining of quotation, please let me know whre to fix my errors.

I don't think anything is wrong. But same comment as before. If you want to present it as a proof reverse the steps, start with, -2ab<=2|a||b| add the squares to both sides, factor and take the square root.

## What is the Triangle Inequality?

The Triangle Inequality states that the sum of any two sides of a triangle must be greater than the third side. In other words, the shortest distance between two points is a straight line.

## Why is the Triangle Inequality important?

The Triangle Inequality is important because it helps determine the validity of a triangle. If the sum of any two sides is less than or equal to the third side, then the triangle is not a valid geometric shape. It is also a fundamental concept in geometry and is used in many mathematical proofs.

## How do you prove the Triangle Inequality?

The Triangle Inequality can be proved using the transitive property of inequalities. This means that if a is less than b and b is less than c, then a is less than c. In the case of the Triangle Inequality, if the sum of two sides is greater than the third side, and the sum of two other sides is also greater than the third side, then the sum of the first two sides must be greater than the sum of the other two sides.

## Can the Triangle Inequality be applied to all triangles?

Yes, the Triangle Inequality applies to all types of triangles, including equilateral, isosceles, and scalene triangles. It is a fundamental property of triangles and is not limited to any specific type or size of triangle.

## How is the Triangle Inequality used in real life?

The Triangle Inequality has many real-life applications, such as in navigation, engineering, and architecture. In navigation, it is used to determine the most direct route between two points. In engineering, it is used to ensure the stability and strength of structures. In architecture, it is used to create aesthetically pleasing and functional designs.

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