Q.452. Trigonometric inequality.

Question 452.
In any triangle ABC,  prove that : sin A sin B sin C  ≤  3√3/8.

Answer 452.
f = sinA sinB sinC
= sinA sinB sin[π - (A+B)]
= sinA sinB sin(A+B)
= sinA sinB (sinA cosB + cosA sinB)
= sin^2 A sinB cosB + sin^2 B sinA cosA

To find the maximum value,
δf/δA = 0
=> 2sinA cosA sinB cosB + sin^2 B * cos2A = 0
=> sin2A cosB + cos2A sinB = 0
=> sin(2A + B) = 0
=> 2A + B = π ... ( 1 ) [cannot be zero or higher multiple of π for traingle ABC]

Similarly,
δf/δB = 0
=> 2B + A = π ... ( 2 )

Solving ( 1 ) and ( 2 ),
A = B = π/3 => C = π/3
=> sinA sinB sinC = [sin(π/3)]^3 = (3√3)/8
This can be shown to be maximum and not minimum by taking some arbitrary values of A, B and C.
Taking A = π/2, B = C = π/4
=> sinA sinB sinC = 1/2
which shows that (3√3)/8 is the maximum value of sinA sinB sinC
=> sinA sinB sinC ≤ (3√3)/8.
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For the benefit of the readers, I reproduce the elegant solution given on page 4-66 of the following link.

Click to open link.

Consider three points P(A, sinA), Q(B, sinB) and R(C, sinC)
on the curve y = sinx such that A + B + C = π
Centroid of ΔPQR = [π/3, (1/3)(sinA + sinB + sinC)]
Therefore, centroid lies on x = π/3 and is inside the triangle,
=> (1/3) (sinA + sinB + sinC) ≤ sin(π/3) = √3/2
=> (sinA sinB sinC)^(1/3) ≤ √3/2 ... [ GM ≤ AM ]
=> sinA sinB sinC ≤ (3√3)/8.

Link to YA!