Minimize $\dfrac{x^2+1}{x^2(1-x)(x+3)}$ on $(0,1)$

Question

One can differentiate to obtain

$$f'(x)=\frac{2(x^2+3)(x^2+x-1)}{x^3(1-x)^2(x+3)^2}.$$

Thus, the unique stationary point is $x=\dfrac{\sqrt{5}-1}{2}$, which is also the minimum point on $(0,1)$. But can we solve by a more elementary method, for example, by inequalities?