Define $h(\mu)=2 \frac{e^x-x-1}{x^2}=2\sum_{k=2}^{\infty}\frac{x^{k-2}}{k!}$ , random variable $X\leq b$ almost surely and $\lambda\in[0, \frac{3}{b}) $. Then $$\mathbb{E}[e^{\lambda(X-\mathbb{E}[X])}]\leq e^{-\lambda\mathbb{E}[X]}(1+\lambda \mathbb{E}[X]+\frac{1}{2}\lambda^2\mathbb{E}[X^2]h(\lambda b) )\leq e^{\frac{\lambda^2\mathbb{E}[X^2]}{2} h(\lambda b)}$$, since $\mathbb{E}[e^{\lambda X} ]= 1+\lambda \mathbb{E}[X]+\frac{1}{2}\lambda^2\mathbb{E}[X^2h(\lambda X)]$ and $\lambda X\leq \lambda b$.
I don't understand how the second inequality is derived. But I know $ e^{-\lambda\mathbb{E}[X]}\leq 1$, so we don't need care about it and $1+\lambda \mathbb{E}[X]+\frac{1}{2}\lambda^2\mathbb{E}[X^2]h(\lambda b)$ is part of some taylor expansion and is smaller then $\mathbb{E}[e^{\lambda X h(\lambda b)}]$, but how should I deal with $e^{\frac{\lambda^2\mathbb{E}[X^2]}{2} h(\lambda b)}=\mathbb{E}[e^{\frac{\lambda^2\mathbb{E}[X^2]}{2} h(\lambda b)}]$ and is it greater then $ \mathbb{E}[e^{\lambda X h(\lambda b)}] $ ?
This is part of the proof of one-sided Bernstein's inequality,in the book https://www.cambridge.org/core/books/abs/highdimensional-statistics/basic-tail-and-concentration-bounds/30AF7B572184787F4C99715838549721