Is there a simpler method to prove absolute inequalities?

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    Absolute Inequalities
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The discussion focuses on proving the absolute inequality |x+y+z| ≤ |x| + |y| + |z| for real numbers. A participant suggests starting with the simpler case of |x+y| ≤ |x| + |y|, which holds true when both terms share the same sign or are opposites. The three-term case can be established by applying the two-term proof iteratively. This method emphasizes a systematic approach to handling absolute values in inequalities.

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imranmeghji
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prove the folowing and state when the inequality holds...

|x+y+z|<=|x|+|y|+|z|

i was thinking that i consider all the possible cases, ie x is positive, y positive, z positive; then the various combinations with negative as well...
is there another shorter method of doing it?
help...
 
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You can start with 2 terms first. |x+y|<=|x|+|y|. (I am assuming you are talking about real numbers). Equal if the same sign, less if opposite. Then the 3 term case can be proven in two steps using the 2 term case.
 

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