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

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 ($c_alphaexp(-|x|^{alpha})$, $c_{alpha}>0$ 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 $alpha=2$, ${Gamma(3/alpha)over 2 Gamma(1/alpha)}={1over 4}$ and $ {Gamma(1/alpha) over alpha}= {alpha^2 sqrt{pi} over 8} $ and $mbox{Var}(xi)= {Gamma(3/alpha)overGamma(1/alpha)}$ when $ xi sim c_alphaexp(-|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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