Given that $\epsilon > 0$, $\delta=\sqrt{\epsilon}$, and $x_1$ and $x_2$ are positive integers such that $x_1+x_2\leq 2k$ and $x_i\geq (1-\delta)k$ for each $i$, where $k$ is a positive integer, show that
$$ 2\cdot 2^{x_1} \cdot 2^{x_2-k/4}\cdot 4^{2\delta k} \leq\frac{(4-2\epsilon)^k}{4\delta k}.$$
The above is equivalent to showing that $\delta k \cdot 2^{3+x_1+x_2-k/4+4\delta k} \leq (4-2\epsilon)^k$. Then
$$\delta k 2^{3+x_1+x_2-k/4+4\delta k}=\delta k \cdot 4^{\frac{3+x_1+x_2-k/4+4\delta k}{2}} \leq \delta k \cdot 4^{\frac{3+2k-k/4+4\delta k}{2}}= \delta k \cdot 8 \cdot (4^{7/8+2\delta})^k.$$.
I am not sure about where to go from here.