I've come across a question that firstly asks to derive the moment-generating function of a random variable $X \sim \mathrm{Normal}(\mu, \sigma^2)$ as $$ M_X(t) = \exp\left(\mu t + \frac{\sigma^2t^2}{2}\right), $$ which is fairly straightforward. Then, it asks to derive the following bound on the probability that $X$ exceeds a certain value: $$ \forall \delta > 0 . \mathbb{P}(X \geq (1+\delta)\mu) \leq \exp\left(-\frac{d^2\mu^2}{2}\right) $$
Now, I make the following attempt: $$ \begin{align} \mathbb{P}(X \geq (1+\delta)\mu ) & \leq \min_{t \in \mathbb{R}}\left( \frac{\exp\left(\mu t + \frac{\sigma^2t^2}{2}\right)}{\exp((1+\delta)\mu t)}\right) \\ \\ & = \min_{t\in \mathbb{R}} \left(\exp\left(\frac{\sigma^2t^2}{2} - \delta t \mu\right)\right) \\ \\ &\stackrel{\mathrm{def}}{\equiv} \min_{t\in \mathbb{R}}(f(t)) \end{align} $$ Now, optimising over $t$ gives: $$ \begin{align} && \frac{\partial f}{\partial t} &= f(t) \cdot (\sigma^2t-\delta \mu)\\ \therefore && \frac{\partial f}{\partial t} = 0 &\implies t= \delta\mu\sigma^{-2} \end{align} $$ Thus, the tightest bound that can be derived using Markov's inequality with the mgf is $$ \forall \delta > 0 . \mathbb{P}(X \geq (1+\delta)\mu) \leq \exp\left(\frac{\delta^2\mu^2}{2\sigma^2}-\frac{\delta^2\mu^2}{\sigma^2}\right) =\exp\left(-\frac{\delta^2\mu^2}{2\sigma^2}\right) $$
This, however, is only a tighter bound than the desired one if $\sigma^2 \leq 1$, which is not specified in the question.
Is this because the question is flawed in not specifying this constraint, or is there an alternative strategy than the Markov bound which may be used to obtain the bound?
EDIT: As a commenter pointed out, the bound the question gives cannot hold in general because we can make the given probability arbitrarily high for fixed $\delta$ by setting $\sigma^2$ arbitrarily close to $\frac{1}{2}$. The question is therefore flawed.