Let $X$ be a non-negative r.v. so that $$ P(X \geq t) \leq C \exp\bigg\{\frac{-t^2/2}{\sigma^2 + bt}\bigg\}$$ for positive $\sigma, b$ and $C\geq 1$. Show that
$$ E[X] \leq 2\sigma (\sqrt{\pi} + \sqrt{\log C}) + 4b(1 + \log C) $$
I know how to usually get concentration bound or bound the deviation from the mean, but no idea how to bound this expectation here. Any help would be really appreciated.
Hold on, this inequality doesn't appear true. Let's take $C = 100, \sigma^2 = 1, b = 10^{-6}$ and suppose the tail-bound were exact. Then \begin{align*} E[X] = 100\int_0^\infty \exp\left(\frac{-t^2}{2(1 + 10^{-6}t)}\right)dt \approx 125.33 \end{align*} Yet \begin{align*} 2\cdot 1 \cdot (\sqrt{\pi}+\sqrt{\log 100}) + 4\cdot 10^{-6}\cdot(1+\log 100) \approx 7.84 \end{align*}