Proof of triangle inequality I understand intuitively that this is true, but Im

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Proof of triangle inequality  I understand intuitively that this is true, but Im

autormtak0w · Accepted Answer

From your definition of the absolute value, establish first |x|=max{x,−x} and ±x≤|x|.   Then you can use   a+b−a−b≤|a|+b≤|a|+|b|,and  ≤|a|−b≤|a|+|b|.

Samantha Brown · Accepted Answer

a2+b2+2|a||b|≥a2+b2+2ab  (|a|+|b|)2≥|a+b|2(∴∀x∈R;x2=|x|2)  ∴|a|+|b|≥|a+b|

nick1337 · Accepted Answer

If a neat algebraic argument does not suggest itself, we can do a crude argument by cases, guided by the examples $a = 7, b = 4, a = - 7, b = - 4, a = 7, b = - 4, a n d a = - 7, b = 4$
If $a \geq 0 a n d b \geq 0$ then $| a + b | = | a | + | b | .$
If $a \leq 0, a n d b \leq -$ , then $| a + b | = - (a + b) = (- a) + (- b) = | a | + | b | .$
Now we need to examine the cases where a is positive and b is negative, or the other way around. Without loss of generality we may assume that $| b | \leq | a | .$
If $a > 0, t h e n | a + b | = | a | - | b |$ . This is $< | a |$ , and in particular $< | a | + | b |$ .
If $a < 0$ , then again $| a + b | = | a | - | b |$

Proof of triangle inequality I understand intuitively that this is true,

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