Absolute Value Proof (Need Help)

Seda
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Homework Statement



Prove that abs(a+b+c) is less than or equal to abs(a)+abs(b)+abs(c)

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The Attempt at a Solution



This makes sense to me that this would always be true, but i just can't seem to figure out how to write it out
 
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Write it as abs((a+b)+c) then use the triangle inequality a couple times.
 
Are you asking to prove |a+b+c| < |a| + |b| + |c| given |a+b| < |a| + |b|? Or do you need to prove |a+b| < |a| + |b| as well?
 
Yes, we are given that the triangle inequality is true, (and we also know how to prove the triangle inequality if that helps.)

But it doesn't seem that I can prove this the same way.

I know abs(a+b) <= abs(a) + abs(b)

So I can very easily get that abs(a+b) + abs(c) <= abs(a) + abs(b) + abs(c)

But how do I a get the left half to what i want?
 
Let d = a+b. |d+c| < |d|+|c| by tri. ineq.

So, |a+b+c| < |a+b|+|c|.

Now use the tri. ineq. on a+b, then add |c| to both sides.
 
Seda said:
Yes, we are given that the triangle inequality is true, (and we also know how to prove the triangle inequality if that helps.)

But it doesn't seem that I can prove this the same way.

I know abs(a+b) <= abs(a) + abs(b)

So I can very easily get that abs(a+b) + abs(c) <= abs(a) + abs(b) + abs(c)

But how do I a get the left half to what i want?

Think of (a+b) as a single number. That is, look at |x+ y| and then set x= a+b, y= c.
 
Wow...it seems so obvious now...

Sometimes I feel like a genius, others I feel stupid as hell, guess what's the case now?

Thanks so much.
 
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