China Mathematical Competition

As $$\begin{array}{rl}f(x)& ={\sin }^{4}x-\sin x\cos x+{\cos }^{4}x\\ & =1-\frac{1}{2}\sin 2x-\frac{1}{2}{\sin }^{2}2x,\end{array}$$ 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 $$\underset{-1\le t\le 1}{\min }g(t)=g(1)=\frac{9}{8}-\frac{1}{2}\times \frac{9}{4}=0,$$ and $$\underset{-1\le t\le 1}{\max }g(t)=g\left(-\frac{1}{2}\right)=\frac{9}{8}-\frac{1}{2}\times 0=\frac{9}{8}.$$ Hence $0\le f(x)\le \frac{9}{8}$.