# Thread: triangle innequality question

1. ## triangle innequality question

the question and prove:
http://i33.tinypic.com/2gvk7k9.jpg

they use the triangle innequality to prove itself
??

2. That's not the triangle inequality that they proved. The proof is fine.

3. Originally Posted by transgalactic
the question and prove:
http://i33.tinypic.com/2gvk7k9.jpg

they use the triangle innequality to prove itself
??
Problem: Prove that $\displaystyle |x|-|y|\le|x-y|$

Proof:

Lemma: $\displaystyle |x+y|\le |x|+|y|$

Proof: Since $\displaystyle \left||x|+|y|\right|=|x|+|y|$ the above is equivanlent to $\displaystyle \sqrt{\left(x+y\right)^2}\le\sqrt{\left(|x|+|y|\ri ght)^2}$. Expanding both sides gives $\displaystyle x^2+2xy+y^2\le|x|^2+2|x||y|+|y|^2$. Noticing that $\displaystyle |x|^2=|x|\cdot|x|=\left|x\cdot x\right|=x^2$ we may rewrite this inequality as $\displaystyle x^2+2xy+y^2\le x^2+2|x||y|+y^2$ or $\displaystyle xy\le |x||y|$ which is trivially true. $\displaystyle \blacksquare$

Particularly the above tells us that $\displaystyle |x|=\left|x-y+y\right|\le |x-y|+|y|$ or equivalently $\displaystyle |x|-|y|\le |x-y|$.

Remark: Since we could have equally as well proven that $\displaystyle |y|-|x|\le|x-y|$ we may actually strengthen this claim to $\displaystyle \bigg||x|-|y|\bigg|\le|x-y|$.

4. your lemma is a half of the triangle innequality law
and the thing i need to prove is its other half.
so we using the law to prove the law itself

??

5. Originally Posted by transgalactic
your lemma is a half of the triangle innequality law
and the thing i need to prove is its other half.
so we using the law to prove the law itself

??
I'm not sure what you mean. All the triangle inequality says is that $\displaystyle \|x+y\|\leq\|x\|+\|y\|$. You can (and should) use this to prove the given problem.

6. ahhhh so what i need to prove is not the triangle innequality law
in calc1 i was told other wise

thanks