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Is my proof of this inequality correct?

  1. Apr 20, 2013 #1
    1. The problem statement, all variables and given/known data

    Prove that |a + b| ≤ |a| + |b|.

    2. Relevant equations

    |a| = √a2

    3. The attempt at a solution

    Since |a| = √a2, then

    |a + b| = √(a + b)2 = √(a2 + 2ab + b2) = √a2 + √b2 + √(2ab) = |a| + |b| + √(2ab).

    And since the square root of a negative number is not defined, then 2ab must be ≥ 0.

    This proves the theorem that |a + b| ≤ |a| + |b|.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Apr 20, 2013 #2

    jbunniii

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    Gold Member

    Alas, this step is incorrect. To see that this is the case, try ##a=b=1##.

    Instead of working with square roots, may I suggest trying to prove this inequality:
    $$(|a+b|)^2 \leq (|a|+|b|)^2$$
    Then argue that this inequality implies the one you want.
     
  4. Apr 20, 2013 #3
    Thanks, jbunniii, for the hint. As you can see, my algebra is a bit rusty. This is what I came up with after your hint.

    To prove that |a + b| ≤ |a| + |b|, we will first attempt to prove that (|a + b|)2 ≤ (|a| + |b|)2.

    Since (|a + b|)2 is equal to (a + b)2, we have

    a2 + b2 + 2ab ≤ (|a| + |b|)2.

    This gives us

    a2 + b2 + 2ab ≤ a2 + b2 + 2|a||b|

    which reduces to

    ab ≤ |a||b|

    which must be true.

    Thus we can take the square root of both sides of the original equation to prove the theorem that |a + b| ≤ |a| + |b|.
     
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