I am trying to check that: $$f(t) = t^2(t-3) + 2 e^{-t/2} + e^{t/2}\left(\sqrt{3} \sin \frac{\sqrt{3}t}{2} + \cos \frac{\sqrt{3} t}{2} \right) - e^{-t/2} \left(\sqrt{3} \sin \frac{\sqrt{3}t}{2} + 3\cos \frac{\sqrt{3} t}{2} \right)\ge 0$$ for $ t \in [0, \infty)$ ( in fact for $t\in \mathbb{R}$). This function equals $6\times$ the inverse Laplace transform of $$\phi(s) =\frac{1}{s^4(1+s+s^2+s^3+s^4+s^5)}= \frac{1}{s^4(1+s)(1-s+s^2)(1+s+s^2)}$$
I don't have a clean proof of the fact. Any feedback would be appreciated. Thank you for your interest!
$\bf{Note:}$ the exponent $5=6-1$ in $\phi$ is chosen since $6$ factors nicely and WA can calculate the inverse Laplace transform. The exponent $4$ is so that we get a positive result ( anything larger would work). We could instead use $7=8-1$, or $11= 12-1$, and an exponent $m$. For $m$ large enough we get again a positive $f(t)$. Here we are looking for $\phi(s)$ that are completely monotone
I tried looking at the Taylor expansion of $f(t)$. One could perhaps prove that is is alternating, so that shows the inequality for $t\le 0$. But we are interested in $t\ge 0$. It seems that the function $\frac{1}{f(t)}$ has a Taylor expansion at $t=0$ with positive coefficients.
The conclude is wrong. Note \begin{eqnarray} f(t) &=& t^2(t-3) + 2 e^{-t/2} + e^{t/2}\left(\sqrt{3} \sin \frac{\sqrt{3}t}{2} + \cos \frac{\sqrt{3} t}{2} \right) - e^{-t/2} \left(\sqrt{3} \sin \frac{\sqrt{3}t}{2} + 3\cos \frac{\sqrt{3} t}{2} \right)\\ &=& t^2(t-3) + 2 e^{-t/2} + 2e^{t/2}\sin \left(\frac{\sqrt{3}t}{2}+\frac{\pi}{3} \right) - e^{-t/2} \left(\sqrt{3} \sin \frac{\sqrt{3}t}{2} + 3\cos \frac{\sqrt{3} t}{2} \right). \end{eqnarray} Let $\frac{\sqrt{3}t_n}{2}+\frac{\pi}{3}=2n\pi-\frac{\pi}{4}$ (where $n$ is a positive integer). Then $$ f(t_n)=t_n^2(t_n-3) + 2 e^{-t_n/2} - \sqrt2e^{t_n/2}-e^{-t_n/2} \left(\sqrt{3} \sin \frac{\sqrt{3}t_n}{2} + 3\cos \frac{\sqrt{3} t_n}{2} \right)$$ and hence $f(t_n)\to-\infty$ as $n\to\infty$.