China Mathematical Competition (Jiangxi)

Since $f(x)$ is defined on $(0,+\infty )$, from $$\{\begin{array}{l}2a^2+a+1=2{\left(a+\frac{1}{4}\right)}^2+\frac{7}{8}>0,\\ 3a^2-4a+1=(3a-1)(a-1)>0,\end{array}$$ 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 $1<a<5$.