My Math Forum Triangle Inequality: Prove Absolute Value Inequality

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 September 1st, 2016, 07:56 AM #1 Newbie   Joined: Sep 2016 From: Singapore Posts: 1 Thanks: 0 Triangle Inequality: Prove Absolute Value Inequality Help Please! The Triangle Inequality Theorem states that ||x|-|y||<|x+y|<|x|+|y| Given that |x-2|<(1/3) Prove 4 < |3x-11| < 6 The solution for x is 5/3 < x < 7/3. But I cannot just say "as x is approaching" these values since using the proof is required. Anyone has an idea on how to manipulate |3x-11| to fit into the theorem?
September 1st, 2016, 06:58 PM   #2
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 Originally Posted by StillAlive Help Please! The Triangle Inequality Theorem states that ||x|-|y||<|x+y|<|x|+|y| Given that |x-2|<(1/3) Prove 4 < |3x-11| < 6 The solution for x is 5/3 < x < 7/3. But I cannot just say "as x is approaching" these values since using the proof is required. Anyone has an idea on how to manipulate |3x-11| to fit into the theorem?
Start with $\displaystyle \dfrac{5}{3} < x < \dfrac{7}{3}$. Multiply by 3 and subtract 11.

 September 1st, 2016, 09:09 PM #3 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,446 Thanks: 2499 Math Focus: Mainly analysis and algebra Alternatively, start with $|x-2|< \frac13 \implies -\frac13 < x-2 < \frac13$. Now use normal algebraic manipulations to transform the $x-2$ term into $3x - 11$.
 September 2nd, 2016, 02:08 PM #4 Senior Member   Joined: Aug 2016 From: morocco Posts: 273 Thanks: 32 you can also start by replacing the second 4<|3x-11|<6 by 4<3x-11<6 or -6<3x-11<-4 which gives 5
September 2nd, 2016, 05:42 PM   #5
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 Originally Posted by StillAlive The Triangle Inequality Theorem states that ||x|-|y||<|x+y|<|x|+|y|
Shouldn't each "<" be "$\small\leqslant$"?

 September 2nd, 2016, 11:45 PM #6 Senior Member   Joined: Aug 2016 From: morocco Posts: 273 Thanks: 32 of course. if x=y=0 we have equalities.

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