For a, b, c the sides of a triangle, prove | a b − b a + b c − c b + c a − a c | &lt; 1I initially thought of changing it in terms of sine using the law of sines, which gives me | sin ⁡ A sin ⁡ B − sin ⁡ B sin ⁡ A + sin ⁡ B sin ⁡ C − sin ⁡ C sin ⁡ B + sin ⁡ C sin ⁡ A − sin ⁡ A sin ⁡ C | &lt; 1However, from there I'm not sure how to proceed. Any tips?

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For   a,   b,   c the sides of a triangle, prove       |          a      b        −          b      a        +          b      c        −          c      b        +          c      a        −          a      c        |    &amp;lt;  1I initially thought of changing it in terms of sine using the law of sines, which gives me      |                  sin        ⁡        A                    sin        ⁡        B              −                  sin        ⁡        B                    sin        ⁡        A              +                  sin        ⁡        B                    sin        ⁡        C              −                  sin        ⁡        C                    sin        ⁡        B              +                  sin        ⁡        C                    sin        ⁡        A              −                  sin        ⁡        A                    sin        ⁡        C              |    &amp;lt;  1However, from there I&#039;m not sure how to proceed. Any tips?

decoratesuw · Accepted Answer

Hint: Using the substitution   a  =  x  +  y  ,  b  =  y  +  z  ,  c  =  z  +  x (if   a  ,  b  ,  c are sides of a triangle then   x  ,  y  ,  z always exist and are positive) we get      |          ∑              c        y        c                    (              a        b            −              b        a            )        |    =      |                  (        x        −        y        )        (        y        −        z        )        (        z        −        x        )                    (        x        +        y        )        (        y        +        z        )        (        z        +        x        )              |    &amp;lt;  1

yasusar0 · Accepted Answer

By an inequality of the triangle       |          ∑              c        y        c                    (              a        b            −              a        c            )        |    =                    |                    ∑                  c          y          c                    (              a        2            c      −              a        2            b      )              |                    a      b      c        =                    ∏                  c          y          c                            |            a      −      b              |                    a      b      c        =      ∏          c      y      c                          |            a      −      b              |              c    &amp;lt;  1

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