I came across the following inequality from these notes. It states that $\frac{e^{-\lambda}}{\sqrt{1-2\lambda}} \le e^{2\lambda^2}$ for $\lambda < \frac{1}{4}$.
Is there a way to show this using calculus? I tried setting $f(\lambda) = e^{2\lambda^2} -\frac{e^{-\lambda}}{\sqrt{1-2\lambda}} $. Since $f(0) = 0$, I tried showing that at least for $\lambda \in (0,\frac{1}{4}), f'(\lambda) > 0$ so that $f(\lambda) > 0$. However, it doesn't seem straightforward to show this.
Are there other alternative approaches to show this inequality?
You can take $$ g(\lambda)=\sqrt{1-2\lambda}e^{2\lambda^2+\lambda} $$ and we want to show that $g(\lambda)\ge 1$ for $\lambda\in (0,1/4)$. Note that $g(0)=1$. Then we use the product rule to calculate $$ g'(\lambda)=\frac{-g(\lambda)}{1-2\lambda}+(4\lambda+1)g(\lambda) $$ and we want to show that this is positive, that is $$ (4\lambda+1)\ge \frac{1}{1-2\lambda} $$ We multiply to clear the denominator, and we get that we need $\lambda$ to satisfy $$ -8\lambda^2+2\lambda\ge 0 $$ which it does for $\lambda\in (0,1/4)$.