Can Absolute Value Inequalities Prove This Expression?

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SUMMARY

The discussion centers on proving the expression \((|a| + |b| + |a - b|)/2 < c\) given the conditions \(|a| < c\) and \(|b| < c\). Participants analyze the inequalities and derive that \(|a + b| < 2c\) and \(|a| + |b| < 2c\). The conclusion emphasizes that if \(a\) and \(b\) are real numbers, the proof can be approached by considering the four possible combinations of signs for \(a\) and \(b\).

PREREQUISITES
  • Understanding of absolute value inequalities
  • Familiarity with algebraic manipulation of inequalities
  • Knowledge of real numbers and their properties
  • Basic skills in mathematical proof techniques
NEXT STEPS
  • Study the properties of absolute values in inequalities
  • Learn about algebraic manipulation techniques for inequalities
  • Explore mathematical proof strategies, particularly for inequalities
  • Investigate the implications of complex numbers in inequality proofs
USEFUL FOR

Students studying algebra, mathematicians interested in inequalities, and educators teaching proof techniques in mathematics.

Andrax
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Homework Statement


we know that |a| < c and |bl < c
prove that : (la+bl + la-bl)/2 < c






The Attempt at a Solution


all I've gotten to so far is this : la+bl < 2c
lal + lbl < 2c
we have : la+bl < lal+lbl
then la+bl < 2c
i need to prove that la-bl < 0.?
also by squaring all i gotten so far is (la+bl + la-bl)/2 < 2c...
 
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Hi Andrax! :smile:
Andrax said:
we know that |a| < c and |bl < c
prove that : (la+bl + la-bl)/2 < c

If they're complex, it's not true.

If they're real, then just try the 4 different ± possiblities, :wink:
 

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