Question

The number of triplets $(x, y, z)$ where $x, y, z \in [0, \pi]$ satisfying the inequlity $(4 + \sin 4x) (2 + \cot^2 y) (1 + \sin^4 z) \leq 12 \sin^2 z$ is

Answer 1

2

Answer 2

The inequality can be rewritten as $(4 + \sin 4x) (2 + \cot^2 y) \leq \frac{12 \sin^2 z}{1 + \sin^4 z}$. The minimum value of the LHS is 6, and the maximum value of the RHS is 6. Thus, equality must hold for both sides, leading to $\sin 4x = -1$, $\cot^2 y = 0$, and $\sin^2 z = 1$. Solving these trigonometric equations for $x, y, z \in [0, \pi]$ yields 2 solutions for $x$, 1 for $y$, and 1 for $z$, resulting in $2 \times 1 \times 1 = 2$ triplets.