, 2014

Let $s$ stand for $\sin t$, $c$ for $\cos t$, and let $x = 2sc = \sin(2t)$. Then
$$\begin{aligned} & (\sin^3 t + \cos^3 t)(\sin t + \cos t)^3 + 16 \sin^3 t \cos^3 t \\ &= (s^3 + c^3)(s + c)^3 + 2x^3 \\ &= (s^2 - sc + c^2)(s + c)^4 + 2x^3 \\ &= (1 - sc)(s^2 + 2sc + c^2)^2 + 2x^3 \\ &= \left(1 - \frac{1}{2}x\right) (1 + x)^2 + 2x^3 = 1 + \frac{3}{2}x + \frac{3}{2}x^3. \end{aligned}$$
Hence, the inequality reduces to proving that
$$1 + 3x \ge 1 + \frac{3}{2}x + \frac{3}{2}x^3 \quad \text{or equivalently} \quad x(1 - x^2) \ge 0,$$
which is true since $0 < x \le 1$. There is equality iff $\sin^2(2t) = x^2 = 1$, equivalently, iff $\cos^2(2t) = 0$ as stated.