Proving Triangle Inequality: How to Justify Your Work | Homework Tips

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The discussion focuses on proving the triangle inequality, specifically the relationship la - bl <= lal + lbl. The initial proof attempts to establish this inequality through algebraic manipulation and squaring both sides. Participants suggest justifying each step by analyzing cases, particularly when dealing with products of real numbers, and highlighting that adding the same quantity to both sides of an inequality preserves the inequality. The conversation emphasizes the importance of clearly explaining each step in the proof process to provide a solid justification. Overall, the proof presented is deemed valid with appropriate justification methods discussed.
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Homework Statement



la - bl <= lal + lbl

Homework Equations





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 don't feel very comfortable justifying inequalities, let alone I am not very good at it.
are there any ways to justify without feeling too awkard? some keywords i need to be using?
please help, i want to learn more
 
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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&gt; 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.
 

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