y = A csc ( A ) + B csc ( B ) + C csc ( C )
For 0 < A , B , C ≤ 2 π , find the maximum value of y .
Notation: csc ( ⋅ ) denotes the cosecant function.
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I was thinking if we could prove it by Jensen's inequality. But we finally reach here, no matter.
x c o s e c ( x ) is an increasing function. in (0,π/2).
Because ,
Consider f ( x ) = x c o s e c ( x ) ,
f ′ ( x ) = c o s e c ( x ) − x c o s e c ( x ) c o t ( x ) ,
f ′ ( x ) = c o s e c ( x ) ( 1 − x c o t ( x ) ) ,
f ′ ( x ) = ( c o s e c ( x ) ( t a n ( x ) − x ) / t a n ( x ) ,
We know that, s i n ( x ) < x < t a n ( x ) .
So , we can conclude that f ′ ( x ) is positive in(0,π/2).
So f ( x ) is increasing function,
So it is maximum at π/2 ,
Therefore, the given equation has maximum at A=B=C=π/2.
So maximum value is 3π/2 = 4.172 .
Awsome solution
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Since A csc A , B csc B and C csc C are identical function, y is maximum when x csc x is maximum. We shall prove that f ( x ) = x csc x = sin x x is an increasing function for 0 < x ≤ 2 π . Therefore, the maximum of f ( x ) , f m a x = f ( 2 π ) = sin 2 π 2 π = 2 π . And the maximum pf y , y m a x = 3 f m a x = 2 3 π ≈ 4 . 7 1 2 .
Proof: Let us prove that f ( x ) = sin x x is an increasing function for 0 < x < 2 π . We need to prove that f ′ ( x ) > 1 .
f ( x ) f ′ ( x ) = sin x x = d x d ( sin x x ) = sin x 1 − sin 2 x x cos x = sin x 1 ( 1 − tan x x ) = sin x 1 ( 1 − x + 3 x 3 + 1 5 2 x 5 + . . . x ) = sin x 1 ( 1 − 1 + 3 x 2 + 1 5 2 x 4 + . . . 1 ) By Maclaurin series Dividing up and down by x Note that 1 + 3 x 2 + 1 5 2 x 4 + . . . 1 < 1
We note that sin x 1 > 0 and 1 − tan x x > 0 , ⟹ f ′ ( x ) > 0 and f ( x ) , an increasing function, for 0 < x < 2 π .