Can Absolute Value Inequalities Prove This Expression?

AI Thread Summary
The discussion revolves around proving the expression (|a+bl + |a-bl)/2 < c given the conditions |a| < c and |bl| < c. Initial attempts suggest that |a+bl| < 2c and |a| + |bl| < 2c, leading to the conclusion that |a+bl| < |a| + |bl|. A key point raised is the need to demonstrate that |a-bl| < 0. The conversation also notes that the proof may not hold if the variables are complex, and suggests exploring the four possible combinations of signs for a comprehensive solution. The focus remains on establishing the inequality under the given constraints.
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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