Prove this generalization of Cauchy-Schwarz: | ( a . b ) c + ( b . c ) a − ( c . a ) b | ≤ | a | | b | | c | If you take c unit perpendicular to a,b, Cauchy-Schwarz follows.I can prove the inequality in a roundabout way. First normalize to unit vectors. Then show that equality holds for 2-dimensional vectors (a bit messy). Then split c=d+e with d ∈ S p a n ( a , b ), e ⊥ d and apply | ( a . b ) d + ( b . d ) a − ( a . d ) b + ( a . b ) e | 2 = | d | 2 + ( a . b ) 2 | e | 2 ≤ | d | 2 + | e | 2 = 1But it's begging for a straightforward proof.

Question

Prove this generalization of Cauchy-Schwarz:       |    (  a  .  b  )  c  +  (  b  .  c  )  a  −  (  c  .  a  )  b      |    ≤      |    a      |        |    b      |        |    c      |  If you take c unit perpendicular to a,b, Cauchy-Schwarz follows.I can prove the inequality in a roundabout way. First normalize to unit vectors. Then show that equality holds for 2-dimensional vectors (a bit messy). Then split c=d+e with   d  ∈      S    p    a    n    (  a  ,  b  ),   e  ⊥  d and apply      |    (  a  .  b  )  d  +  (  b  .  d  )  a  −  (  a  .  d  )  b  +  (  a  .  b  )  e            |        2    =      |    d            |        2    +  (  a  .  b      )    2        |    e            |        2    ≤      |    d            |        2    +      |    e            |        2    =  1But it&#039;s begging for a straightforward proof.

Calvin Maddox · Accepted Answer

Consider the Gramian matrix for the vectors a,b,c:  G  =      (                            ‖          a                      ‖            2                                    a          ⋅          b                          a          ⋅          c                                      b          ⋅          a                          ‖          b                      ‖            2                                    b          ⋅          c                                      c          ⋅          a                          c          ⋅          b                          ‖          c                      ‖            2                                )    .Recall that G is positive-semidefinite, and positive-definite if and only if a,b,c are linearly independent. Either way, we have   det  G  ≥  0, hence                    0                            ≤        ‖        a                  ‖          2                (        ‖        b                  ‖          2                ‖        c                  ‖          2                −        (        b        ⋅        c                  )          2                )        −        (        a        ⋅        b        )        (        ‖        c                  ‖          2                (        b        ⋅        a        )        −        (        b        ⋅        c        )        (        c        ⋅        a        )        )                                          +        (        a        ⋅        c        )        (        (        b        ⋅        a        )        (        c        ⋅        b        )        −        ‖        b                  ‖          2                (        c        ⋅        a        )        )                                          =        ‖        a                  ‖          2                ‖        b                  ‖          2                ‖        c                  ‖          2                −        ‖        a                  ‖          2                (        b        ⋅        c                  )          2                −        ‖        b                  ‖          2                (        a        ⋅        c                  )          2                −        ‖        c                  ‖          2                (        a        ⋅        b                  )          2                +        2        (        a        ⋅        b        )        (        b        ⋅        c        )        (        a        ⋅        c        )                                          =        ‖        a                  ‖          2                ‖        b                  ‖          2                ‖        c                  ‖          2                −        ‖        (        a        ⋅        b        )        c        +        (        b        ⋅        c        )        a        −        (        c        ⋅        a        )        b                  ‖          2                .            Therefore, the inequality holds, and equality holds if and only if a,b,c are linearly dependent.

Prove this generalization of Cauchy-Schwarz: |(a \cdot b)c+(b \cdot c)a-(c \cdot a)b| <= |a| |b| |c|

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