China Mathematical Competition (Jiangxi)

Since $f(x)$ is defined on $(0, +\infty)$, from
$$\begin{cases} 2a^2 + a + 1 = 2\left(a+\frac{1}{4}\right)^2 + \frac{7}{8} > 0, \\ 3a^2 - 4a + 1 = (3a-1)(a-1) > 0, \end{cases}$$
we get $\qquad$ $a > 1$ or $a < \frac{1}{3}$. $\qquad$ (1)

Since $f(x)$ is a decreasing function on $(0, +\infty)$, so
$$2a^2 + a + 1 > 3a^2 - 4a + 1 \Rightarrow a^2 - 5a < 0.$$
Thus $0 < a < 5$. Combining this with (1), we have $0 < a < \frac{1}{3}$ or $1 < a < 5$.