Prove that $\sum_{cyc}a^4b \ge \sum_{cyc} a^2$

First of all, you homogenised incorrectly. Homogenising instead gives $$\sum_{\text{cyc}} a^4b\geq \sum_{\text{cyc}} a^{11/4}b^{3/4}c^{3/4}d^{3/4}$$ Now by weighted AM-GM $$\frac{133}{204}a^4b+\frac{5}{204}b^4c+\frac{37}{204}c^4d+\frac{29}{204}d^4a\geq a^{11/4}b^{3/4}c^{3/4}d^{3/4}$$ and cyclically adding this inequality and analogous ones yields the result.

EDIT: Here's how to come up with those mysterious coefficients. We want to apply weighted AM-GM such that $$\alpha\cdot a^4b+\beta\cdot b^4c+\gamma\cdot c^4d+\delta\cdot d^4a\geq a^{11/4}b^{3/4}c^{3/4}d^{3/4}$$ for some $\alpha, \beta, \gamma, \delta$ and then cyclically add up this inequality. To make this work, we need \begin{equation*} \begin{split} 4\alpha+\delta&=\frac{11}{4} \\ 4\beta+\alpha&=\frac{3}{4} \\ 4\gamma+\beta&=\frac{3}{4} \\ 4\delta+\gamma&=\frac{3}{4}. \end{split} \end{equation*} Thus, we solve this linear system and obtain the coefficients from the above solution.