Solution:
Note that $2 - \log_{4} x \geq 0 \Longrightarrow 0 < x \leq 16$. Let $t = \sqrt{2 - \log_{4} x}$. Then
$$\log_{4} x = 2 - t^2 \Longrightarrow \frac{\log_{1/2} x}{\log_{1/2} 4} = 2 - t^2 \Longrightarrow \log_{1/2} x = 2 t^2 - 4$$
Substituting back to the given inequality, we have
$$2 t^2 - 4 - t + 1 \leq 0 \Longrightarrow 2 t^2 - t - 3 \leq 0 \Longrightarrow -1 \leq t \leq \frac{3}{2}$$
Since $t = \sqrt{2 - \log_{4} x}$, this means that $0 \leq \sqrt{2 - \log_{4} x} \leq \frac{3}{2}$, which has solution $\frac{1}{\sqrt{2}} \leq x \leq 16$.