Just read this from an example that if a random variable $X\sim\mathbf{N}(0,1)$, then $\mathbb{E}(e^{\lambda(X^2-1)}) = \frac{e^{-\lambda}}{\sqrt{1-2\lambda}} \leq e^{2\lambda^2}$ for all $|\lambda|<\frac{1}{4}$.
But how is the upper bound of $\frac{e^{-\lambda}}{\sqrt{1-2\lambda}}$ found?
We have that by $2\lambda =x$ $$ \frac{e^{-\lambda}}{\sqrt{1-2\lambda}} \leq e^{2\lambda^2} \iff f(x)=x+x^2+\log (1-x)\ge 0$$
with $f(0)=0$ and $$f'(x)=\frac{2x^2-x}{x-1} \begin{cases} < 0 \;\text{for}\;x<0\\\\\ge 0 \;\text{for}\;0\le x\le \frac12 \end{cases}$$
which implies that $f(x)\ge 0$ for $|x|\le \frac12$.