With x and y two vectors, the triangle inequality can be written as: (A) | x | − | y | ≤ | x + y | ≤ | x | + | y | which can be broken down into a system of inequalities (labelled as (1) and (2)): { | x | − | y | ≤ | x + y | | x + y | ≤ | x | + | y | then, with the assumption of | y | &gt; 0, (1)-(2) yields: (B) | x | ≤ | x + y | But is this in fact true? Is (B) a direct consequence of (A) when | y | &gt; 0? or is (B) is invalid because (1)-(2) is not allowed as it is not obvious if the subtraction preserves the inequality?

Question

With   x and   y two vectors, the triangle inequality can be written as:                    (A)                          |              x              |        −              |              y              |        ≤              |              x        +              y              |        ≤              |              x              |        +              |              y              |            which can be broken down into a system of inequalities (labelled as (1) and (2)):      {                                    |                  x                  |          −                  |                  y                  |          ≤                  |                  x          +                  y                  |                                              |                  x          +                  y                  |          ≤                  |                  x                  |          +                  |                  y                  |                        then, with the assumption of   |  y  |  &amp;gt;  0, (1)-(2) yields:                    (B)                          |              x              |        ≤              |              x        +              y              |            But is this in fact true? Is (B) a direct consequence of (A) when   |  y  |  &amp;gt;  0? or is (B) is invalid because (1)-(2) is not allowed as it is not obvious if the subtraction preserves the inequality?

Braxton Gallagher · Accepted Answer

No, we cannot subtract inequalities in the following way:  a  ≤  b    c  ≤  d      ⟹    a  −  c  ≤  b  −  dFor example, take   a  =  3  ,  b  =  4  ,  c  =  1  ,  d  =  3 to see that it&#039;s wrong. I think this is what you did to obtain what you have. So  |  x  |  ≤  |  x  +  y  |is wrong. A counterexample in         R    2  :   x  =  (  1  ,  0  )  ,  y  =  (  −  1  ,  0  )

Edith Mayer · Accepted Answer

(B) does not follow from (1)-(2). For a very simple counter-example, work in the 1-dimensional space of real numbers and take x=3, y=-2.

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