Problem. Let $X_1, . . . , X_n$ be independent random variables such that $\mathbf{E}(X_k) = 0$ and $|X_k| \leq 1$ for $k = 1, . . . , n$. Let $X = X_1 + . . . + X_n$. Prove the following strengthening of the Bernstein inequality: $$P(X \geq \alpha n) \leq \left(\frac{1}{(1 − \alpha)^{1−\alpha}\cdot(1 + \alpha)^{1+\alpha}}\right)^{n/2} \quad \text{ for all } 0 \leq \alpha < 1.$$
Hint: Follow the proof of the Bernstein inequality, only do not use that $\frac{e^t + e^{−t}}{2} \leq e^{\frac{t^2}{2}}$
attempt: Following Probability and random processes by grimmett.
Say $\mathbf{P}(X_k)=p$. For small positive values of $\epsilon$, $$\mathbf{P}\left(\frac{1}{n}X_n\geq p +\epsilon\right)=\Sigma_{k \geq n(p+\epsilon)} \mathbf{P}(X_n=k) $$ hence $$\mathbf{P}(X_n=k)=\Sigma_{k=m}^n\binom{n}{k}p^k(1-p)^{n-k}$$ where $m=[n(p+\epsilon)]$ the least integer not less than $n(p+\epsilon)$. Let $\lambda>0$ and note that $\exp(\lambda k) \geq \exp(\lambda n (p+\epsilon))$ if $k \geq m$. Allowing $q=1-p.$ We have:
$$\begin{align} \mathbf{P}\left(\frac{1}{n}X_n \geq p+ \epsilon\right) \leq & \Sigma_{k=m}^n \exp(\lambda[k-n(p+\epsilon)]) \binom{n}{k} p^kq^{n-k} \\ \leq & \exp{-\lambda n \epsilon}\Sigma_{k=0}^n \binom{n}{k} p\exp{(\lambda q)}^k(q\exp{(-\lambda p)})^{n-k} \\ =& \exp{(-\lambda n \epsilon)}(p\exp{(\lambda q)}+q\exp{(-\lambda p)})^n\end{align}$$ By the binomial theorem. By the fact that $e^x \leq x+e^{x^2}$ for $x \in \mathbb{R}$ we get:
$$\begin{align} \mathbf{P}\left(\frac{1}{n} X_n \geq p +\epsilon\right) \leq & \exp{(-\lambda n \epsilon)}[pe^{\lambda^2q^2}+qe^{\lambda^2 p^2}]^n \\ \leq & e^{\lambda^2 n-\lambda n \epsilon}.\end{align}$$
We can pick $\lambda$ to minimize the RHS, $\lambda =\frac{1}{2}\epsilon$, to get:
$$\mathbf{P}\left(\frac{1}{n}X_n \geq p +\epsilon\right)\leq e^{-\frac{1}{4}n\epsilon^2} \quad \text{ for } \epsilon >0 $$
Not sure how to get the inequality asked for.