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Jameson Lucero 2022-07-01 Answered
If a , b , c ( 0 , π 2 ) ,, Then prove that sin ( a + b + c ) sin a + sin b + sin c < 1
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Answers (2)

Jayvion Tyler
Answered 2022-07-02 Author has 23 answers
sin ( a ) + sin ( b ) > sin ( a + b ) if ( a , b ) ( 0 , π )
sin ( a + b + c ) <= sin ( a ) + sin ( b + c ) < sin ( a ) + sin ( b ) + sin ( c )

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pipantasi4
Answered 2022-07-03 Author has 6 answers
Using
sin ( a + b + c ) sin a sin b sin c
= 2 cos ( 2 a + b + c 2 ) sin ( b + c 2 ) 2 sin ( b + c 2 ) cos ( b c 2 )
So
= 2 sin ( b + c 2 ) [ cos ( 2 a + b + c 2 ) cos ( b c 2 ) ]
= 4 sin ( a + b 2 ) sin ( b + c 2 ) sin ( a + c 2 ) < 0 ,
Bcz given a , b , c ( 0 , π 2 ) . So we get a + b 2 , b + c 2 , c + a 2 ( 0 , π 2 )
So we get
sin ( a + b + c ) < sin a + sin b + sin c sin ( a + b + c ) sin a + sin b + sin c < 1

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