I am trying to solve a problem involving variational approximation, where the task is to calculate a value $C$ such that
$$C > \frac{\int_{-\infty}^{\infty} |f'(y)|^2 dy}{\int_{-\infty}^{\infty} \frac{1}{y^2} |f(y)|^2 dy}$$
for a given function $f$ that takes the form $f = \frac{y}{y^{2} + \alpha^{2}}$, where $\alpha$ is a variational parameter. I have done simple plug and chug in the expression for $C$ above and got the relation that $C > \frac{1}{2}$. However, when I do some variational approach, I got $C > 1$, which I have done in this fashion:
Let $I_1$ be the integral expression in the numerator and $I_2$ be the expression for the denominator so that
$$C > \frac{I_1}{I_2}$$
Minimizing with respect to $\alpha$, I get
$$\frac{d}{d\alpha} \left(\frac{I_1}{I_2}\right) = 0$$
$$\frac{I_2 \frac{dI_{1}}{d\alpha} - I_1 \frac{dI_2}{d\alpha}}{I_2 ^{2}} = 0$$
which means that
$$I_2 \frac{dI_{1}}{d\alpha} - I_1 \frac{dI_2}{d\alpha} = 0$$
and can be written as
$$I_2 I_1' -I_1I_2' = 0$$
where the prime represents derivative with respect to $\alpha$. This can be arranged in a way that
$$\frac{I'_{1}}{I_1} = \frac{I'_{2}}{I_2}$$
and after integrating both sides, yields the relation $\ln(I_1) = \ln(I_2)$, and thus $I_1 = I_2$. If this is the case, then we get $C>1$. But that is not just right, I think.
Is there something wrong with the math right here? The thing is, I should get $1/2$ as well when minimizing with respect to $\alpha$. On another note, how can I extend this to functions with two variational parameters $\alpha$ and $\beta$, like for example $f(y) = \frac{y}{(y^2+\alpha^{2})^{\beta}}$?