Since $t^{\alpha n}\le t^k$ for $k\le\alpha n$ and $t\in[0,1]$, we have the bound
$$
\begin{align}
t^{\alpha n}\sum_{k=0}^{\alpha n}\binom{n}{k}
&\le\sum_{k=0}^n\binom{n}{k}t^k\\
&=(1+t)^n
\end{align}
$$
Therefore,
$$
\begin{align}
\frac1{2^n}\sum_{k=0}^{\alpha n}\binom{n}{k}
&\le\left(\frac{1+t}{2t^\alpha}\right)^n\\
&=\left(\frac1{2\alpha^\alpha(1-\alpha)^{1-\alpha}}\right)^n
\end{align}
$$
when $t=\frac{\alpha}{1-\alpha}$. Note that to ensure $t\in[0,1]$, we need $\alpha\in[0,1/2]$.
