I am new to proofs and would like some help understanding how to prove the following abs inequality.
$$| -x-y | \leq |x| + |y|.$$
I think I should take out the negative in the left absolute value function.? Then prove for the triangle inequailty.
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Yes. You are right. – Jun 02 '15 at 05:27
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So, the left side would be |(-1)|*|x+y| => |x+y| then i just prove the triangle inequailty. – Arjun Dhiman Jun 02 '15 at 05:28
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Yes. $\mid (-1) \mid$ equals $1$, so you get equality in your above comment. Then indeed use triangle inequality. – Jun 02 '15 at 05:29
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yes, but after removing the negative do i prove |x+y| <= |x| + |y| because that is the triangle inequality? – Arjun Dhiman Jun 02 '15 at 05:31
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Exactly. Do you also want to prove the triangle inequality ? – Jun 02 '15 at 05:32
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if you dont mind can you show me the proof for this inequality, i am a little confused on what to do. – Arjun Dhiman Jun 02 '15 at 05:33
2 Answers
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$$|-x-y| \leq |x| + |y|\\ |-1(x+y)| \leq |x| + |y|\\ |x+y||-1|\leq |x| + |y|\\ 1|x+y|\leq |x| + |y|\\ |x+y|\leq |x| + |y| $$ I think that you can take it from here as long as you use the fact that $|x| = \max(x, -x)$
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We have $\mid -x-y\mid = \mid -(x+y) \mid = \mid x+y \mid \leq \mid x \mid + \mid y \mid$. For the least step we use the triangle inequality. Supposing you work in $\mathbb R$ look here for a proof: Proof of triangle inequality.
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oh wow, thanks guys. This was really straight forward. get rid of negative, then prove triangle ineqauilty. – Arjun Dhiman Jun 02 '15 at 05:37