Let $X$ be a non-negative r.v., prove that \begin{equation} \inf_{k\in\mathbb{Z}_+} \frac{\mathbb{E}[X^k]}{t^k}\leqslant \inf_{\lambda\geqslant 0}\frac{\mathbb{E}[e^{\lambda X}]}{e^{\lambda t}},\;\forall t>0. \end{equation}
I was given the above problem during a class of the advanced statistics course in my university. I tried attempting it as follows:
First, let us rewrite the exponentials on the RHS of the above equation using the Maclaurin expansion: $$ e^{\lambda X} = \sum_{k=0}^{\infty} \frac{(\lambda X)^k}{k!}, \quad e^{\lambda t} = \sum_{k=0}^{\infty} \frac{(\lambda t)^k}{k!}. $$ $X$ being non-negative allows us to expand the expectation of $e^{\lambda X}$ as follows: $$ \mathbb{E}[e^{\lambda X}] = \mathbb{E}\left[\sum_{k=0}^{\infty} \frac{(\lambda X)^k}{k!}\right] = \sum_{k=0}^{\infty} \frac{\lambda^k\mathbb{E}X^k}{k!}. $$ Now, we can write: $$ \frac{\mathbb{E}e^{\lambda X}}{e^{\lambda t}} = \frac{\sum_{k=0}^{\infty} \frac{\lambda^k\mathbb{E}X^k}{k!}}{\sum_{k=0}^{\infty} \frac{(\lambda t)^k}{k!}} = \frac{\sum_{k=0}^\infty c_\lambda(k)\mathbb{E}X^k}{\sum_{k=0}^\infty c_\lambda(k)t^k},\;\text{where } c_\lambda(k)=\lambda^k/k! $$
But then I got completely stuck. I can see that now we have $\mathbb{E}[X^k]$ in the numerator-placed series and $t^k$ in the denominator-placed series, both are dynamically scaled by $c_\lambda(k)$, and I don't see how we can relate them. Eventually, I thought that taking infimum over $\lambda$ can be the key, as we will minimize the fraction only over the $\lambda$-dependent $c_\lambda(k)$, but as I see that doesn't really help me much.
Any help or advice will be greatly appreciated.
Thank you in advance!
You have already noted that $$\mathbb{E}[e^{\lambda X}] = \sum_{k\ge 0} \frac{\lambda^k \mathbb{E}X^k}{k!}$$ which can further be written as $$\mathbb{E}[e^{\lambda X}]=\sum_{k\ge 0} \frac{\mathbb{E}X^k}{t^k} \frac{(\lambda t)^k}{k!} \ge \left(\inf_{k\in \mathbb{Z}_+} \frac{\mathbb{E}X^k}{t^k} \right)\sum_{k\ge 0} \frac{(\lambda t)^k}{k!} =\left(\inf_{k\in \mathbb{Z}_+} \frac{\mathbb{E}X^k}{t^k} \right)e^{\lambda t} $$ so dividing through by $e^{\lambda t}$ and taking an infimum over $\lambda \ge 0$ gives the result.