the question and prove:
http://i33.tinypic.com/2gvk7k9.jpg
they use the triangle innequality to prove itself
??
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 $\displaystyle \bigg||x|-|y|\bigg|\le|x-y|$.