Prove $\int_{0}^{\infty} \cos(\omega x) \exp(-x^{\alpha}) \, {\rm d} x \ge {\alpha^2 \sqrt{\pi} \over 8} \exp \left( -\frac{\omega^2}{4} \right)$

Question

I would like to prove that

$$\int_{0}^{\infty} \cos(\omega x) \exp(-x^{\alpha}) \, {\rm d} x \ge {\alpha^2 \sqrt{\pi} \over 8} \exp \left( -\frac{\omega^2}{4} \right)$$

for any $\omega > 0$ and $1 < \alpha < 2$.

Here is some research effort. By using the representation of the generalized Gaussian density ($c_\alpha\exp(-|x|^{\alpha})$, $c_{\alpha}>0$ normalizing constant) as Gaussian mixture, I can prove that $$\int_{0}^{\infty} \cos(\omega x) \exp(-x^{\alpha}) \, {\rm d} x \ge {\Gamma(1/\alpha) \over \alpha} \exp \left( -\frac{\omega^2}{2} {\Gamma(3/\alpha)\over \Gamma(1/\alpha)} \right).$$

When $\alpha=2$, ${\Gamma(3/\alpha)\over 2 \Gamma(1/\alpha)}={1\over 4}$ and $ {\Gamma(1/\alpha) \over \alpha}= {\alpha^2 \sqrt{\pi} \over 8} $ and $\mbox{Var}(\xi)= {\Gamma(3/\alpha)\over\Gamma(1/\alpha)}$ when $ \xi \sim c_\alpha\exp(-|x|^{\alpha})$.

Both ${\Gamma(3/\alpha)\over 2 \Gamma(1/\alpha)}$ and $ {\Gamma(1/\alpha) \over \alpha}$ are decreasing in $\alpha \in (1,2)$. I need the result as originally stated as the coefficient in front of $\omega$ in the exponent has to be free from $\alpha$. All of my attempts are tied to probablisitic arguments and hence I fail to get rid of ${\Gamma(3/\alpha)\over 2 \Gamma(1/\alpha)}$ coefficient.

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