shobber wrote: In acute triangle $ ABC$, show that: $ \sin^3{A}\cos^2{(B - C)} + \sin^3{B}\cos^2{(C - A)} + \sin^3{C}\cos^2{(A - B)} \leq 3\sin{A} \sin{B} \sin{C}$ and find out when the equality holds. We only need to prove the equality: $ \sum_{cyc}sin^3Acos(B - C) = 3sinAsinBsinC$ $ 4sin^3Acos(B - C) = 4sin^2(B + C)sin(B - C) = 2sin^2A(sin2B + sin2C) = (1 - cosA)(sin2B + sin2C) = sin2B + sin2C - sin2Bcos2A - sin2Ccos2A$ It follows that: $ \sum_{cyc}4sin^3Acos(B - C) = \sum_{cyc}(sin2B + sin2C) - \sum_{cyc}sin2Bcos2A - \sum_{cyc}sin2Ccos2A$ $ = 2\sum_{cyc}sin2A - \sum_{cyc}(sin2Acos2B + cos2Asin2B)$ $ = 2\sum_{cyc}sin2A + \sum_{cyc}sin2C$ $ = 3(sin2A + sin2B + sin2C)$ $ = 12sinAsinBsinC$ The equality occurs iff $ cos(A - B) = cos(B - C) = cos(C - A) = 1$<=>ABC is an equilateral triangle

Thank you very much. Nice solution

We will prove the miraculous identity $$\sum_{cyc} \sin^3 A \cos (B-C) = 3 \sin A \sin B \sin C.\qquad (1)$$This would clearly imply the desired inequality. First of all, applying the Cosine Subtraction Identity and dividing both sides by $\sin A \sin B \sin C$, we see that $(1)$ is equivalent to: $$\sum_{cyc} \sin^2 A (\cot B \cot C + 1) = 3,$$which in turn is equivalent to $$\sum_{cyc} \sin^2 A \cot B \cot C = \sum_{cyc} \cos^2 A$$after a subtraction. After multiplying by $\tan A \tan B \tan C$, we need only to show that: $$\tan A \tan B \tan C \sum_{cyc} \cos^2 A = \sum_{cyc} \tan^3 A \cos^2 A. \qquad (2)$$Now, set $x, y, z$ to be $\tan A, \tan B, \tan C$ respectively. Using the identity $\cos^2 \theta = \frac{1}{1+ \tan^2 \theta}$ for all angles $\theta$, we can rewrite $(2)$ as: $$xyz \sum_{cyc} \frac{1}{1+x^2} = \sum_{cyc} \frac{x^3}{1+x^2}. \qquad(3)$$Now, we cite the well-known identity that $x+y+z = xyz$. With this in mind, we know that: $$\frac{1}{1+x^2} = \frac{x+y+z}{xyz + x^2(x+y+z)} = \frac{x+y+z}{x(x+y)(x+z)},$$with similar relations for $y, z.$ Plugging this relations into $(3)$, we now want: $$xyz \sum_{cyc} \frac{x+y+z}{x(x+y)(x+z)} = \sum_{cyc} \frac{x^3(x+y+z)}{x(x+y)(x+z)}. \qquad(4).$$After multiplying through by $(x+y)(y+z)(z+x),$ it's easy to find that $(4)$ is simply equivalent to: $$\sum_{cyc} (x+y+z)yz(y+z) = \sum_{cyc} (x+y+z) x^2(y+z),$$which is easily verified to be true. We are now done. $\square$