How do I prove this trigonometric inequality?If A , B , C ∈ ( 0 , π 2 ). Then prove that sin ⁡ ( A + B + C ) sin ⁡ ( A ) + sin ⁡ ( B ) + sin ⁡ ( C ) &lt; 1

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How do I prove this trigonometric inequality?If   A  ,  B  ,  C  ∈  (  0  ,      π    2    ). Then prove that            sin      ⁡      (      A      +      B      +      C      )              sin      ⁡      (      A      )      +      sin      ⁡      (      B      )      +      sin      ⁡      (      C      )        &amp;lt;  1

Triston Donaldson · Accepted Answer

If   A  +  B  +  C  ≥  π our inequality is obviously true.Let   A  +  B  +  C  &amp;lt;  π and   A  ≥  B  ≥  CSince   (  A  +  B  +  C  ,  0  ,  0  )  ≻  (  A  ,  B  ,  C  ) and   sin is a concave function on   [  0  ,  π  ]we obtain:  sin  ⁡  (  A  +  B  +  C  )  +  sin  ⁡  0  +  sin  ⁡  0  ≤  sin  ⁡      A    +  sin  ⁡      B    +  sin  ⁡      C  and we are done!

Keenan Conway · Accepted Answer

Expand using the sum formula for sine:                    sin        ⁡        (        A        +        B        +        C        )                            =        sin        ⁡        (        A        +        B        )        cos        ⁡        (        C        )        +        sin        ⁡        (        C        )        cos        ⁡        (        A        +        B        )                                          =        sin        ⁡        (        A        )        cos        ⁡        (        B        )        cos        ⁡        (        C        )        +        sin        ⁡        (        B        )        cos        ⁡        (        A        )        cos        ⁡        (        C        )        +        sin        ⁡        (        C        )        cos        ⁡        (        A        +        B        )            Now since   A  ,  B  ,  C  ∈      (    0    ,          π      2        )  , we know   max  {  cos  ⁡  (  A  )  ,  cos  ⁡  (  B  )  ,  cos  ⁡  (  C  )  ,  cos  ⁡  (  A  +  B  )  }  &amp;lt;  1, and therefore  sin  ⁡  (  A  )  cos  ⁡  (  B  )  cos  ⁡  (  C  )  +  sin  ⁡  (  B  )  cos  ⁡  (  A  )  cos  ⁡  (  C  )  +  sin  ⁡  (  C  )  cos  ⁡  (  A  +  B  )  &amp;lt;  sin  ⁡  (  A  )  +  sin  ⁡  (  B  )  +  sin  ⁡  (  C  )  .Combining what we have so far, we see that   sin  ⁡  (  A  +  B  +  C  )  &amp;lt;  sin  ⁡  (  A  )  +  sin  ⁡  (  B  )  +  sin  ⁡  (  C  ), and since the left hand side cannot be zero, we can divide by it to obtain the result.

If A,B,C in (0,pi/2). Then prove that (sin(A+B+C))/(sin(A)+sin(B)+sin(C)<1

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