China Mathematical Competition

As
$$\begin{aligned} f(x) &= \sin^4 x - \sin x \cos x + \cos^4 x \\ &= 1 - \frac{1}{2} \sin 2x - \frac{1}{2} \sin^2 2x, \end{aligned}$$
we define $t = \sin 2x$, then
$$f(x) = g(t) = 1 - \frac{1}{2}t - \frac{1}{2}t^2 = \frac{9}{8} - \frac{1}{2}\left(t + \frac{1}{2}\right)^2.$$
So we have
$$\min_{-1 \le t \le 1} g(t) = g(1) = \frac{9}{8} - \frac{1}{2} \times \frac{9}{4} = 0,$$
and
$$\max_{-1 \le t \le 1} g(t) = g\left(-\frac{1}{2}\right) = \frac{9}{8} - \frac{1}{2} \times 0 = \frac{9}{8}.$$
Hence $0 \le f(x) \le \frac{9}{8}$.