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September 1st, 2016, 08:56 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 xy<x+y<x+y Given that x2<(1/3) Prove 4 < 3x11 < 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 3x11 to fit into the theorem? 
September 1st, 2016, 07:58 PM  #2  
Senior Member Joined: Feb 2010 Posts: 632 Thanks: 103  Quote:
 
September 1st, 2016, 10:09 PM  #3 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,031 Thanks: 2342 Math Focus: Mainly analysis and algebra 
Alternatively, start with $x2< \frac13 \implies \frac13 < x2 < \frac13$. Now use normal algebraic manipulations to transform the $x2$ term into $3x  11$. 
September 2nd, 2016, 03:08 PM  #4 
Senior Member Joined: Aug 2016 From: morocco Posts: 273 Thanks: 32 
you can also start by replacing the second 4<3x11<6 by 4<3x11<6 or 6<3x11<4 which gives 5<x<17/3 or 5/3<x<7/3. now let's return to the first x2<1/3 gives 5/3<x<7/3. conclusion if 5/3<x<7/3 then 5<x<17/3 or 5/3<x<7/3. 
September 2nd, 2016, 06:42 PM  #5 
Global Moderator Joined: Dec 2006 Posts: 18,145 Thanks: 1418  
September 3rd, 2016, 12:45 AM  #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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absolute, calculus, inequality, prove, triangle 
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