For a cyclic sum the majorization is not enough.
For example, for non-negative variables $(3,2,0)\succ(3,1,1)$ but the inequality
$$a^3b^2+b^3c^2+c^3a^2\geq a^3bc+b^3ac+c^3ab$$ is wrong.
Try $a\rightarrow+\infty$ and $b^2-bc<0$.
By the way, there is the following way to prove of the Murhead's type cyclic inequalities.
We'll prove that $$\sum_{cyc}a^5b^2\geq \sum_{cyc}a^4b^2c$$ for non-negative variables.
Indeed, by AM-GM $$\sum_{cyc}a^5b^2=\frac{1}{19}\sum_{cyc}(14a^5b^2+2b^5c^2+3c^5a^2)\geq\sum_{cyc}\sqrt[19]{\left(a^5b^2\right)^{14}\left(b^5c^2\right)^2\left(c^5a^2\right)^3}=\sum_{cyc}a^4b^2c.$$
Vasile Cirtoaje was first, which proved that if this way does not work, so the inequality is wrong.
For example, we'll try to prove that $$\sum_{cyc}a^3b^2\geq \sum_{cyc}a^3bc$$ by this way.
We'll try to find values of $\alpha$, $\beta$ and $\gamma$ such that $\alpha+\beta+\gamma=1$ and the inequality
$$\alpha a^3b^2+\beta b^3c^2+\gamma c^3a^2\geq a^3bc$$ would be true by AM-GM.
Indeed, by AM-GM $$\alpha a^3b^2+\beta b^3c^2+\gamma c^3a^2\geq a^{3\alpha+2\gamma}b^{2\alpha+3\beta}c^{2\beta+3\gamma}$$ and we obtain the following system:
$3\alpha+2\gamma=3,$ $2\alpha+3\beta=1$ and $\alpha+\beta+\gamma=1$, which gives $$(\alpha,\beta,\gamma)=\left(\frac{5}{7},-\frac{1}{7},\frac{3}{7}\right)$$ and since $-\frac{1}{7}<0$, this way does not give a proof, which says that the inequality is wrong.
Finding an answer by weighted AM-GM and Muirhead is very directly linked, though it may not simplify the work needed in practice, you'll have to solve a set of simultaneous linear equations in either case. Majorisation $x \succ y$ (ie. Muirhead) is equivalent to $y$ being a convex combination of $x$ and its permutations. Using your example, this means:
$$(4,0,0) \succ (2, 1, 1) \iff w_1(4,0,0)+w_2(0, 4, 0)+w_3(0, 0, 4)=(2, 1, 1)$$
for some non-negative weights $w_i$ s.t. $\sum_i w_i=1$. In fact finding these $w_i$ will give you AM-GM coefficients to use immediately. In this case, it is easily seen the weights $w = (\frac12, \frac14, \frac14)$ does the job, so on cyclically summing
$$\frac12a^4+\frac14b^4+\frac14c^4 \geqslant a^2bc$$
you get an equivalent AM-GM proof.
Best Answer
Muirhead doesn't work, because it is wrong in general for cyclical sums. However, my suggested approach in the comments does work:
For any $\alpha,\beta,\gamma,\delta\ge0$ with $\alpha+\beta+\gamma+\delta>0$,
$$\alpha a^4 b c+\beta b^4 c d+\gamma c^4 d a+\delta d^4 a b\ge(\alpha+\beta+\gamma+\delta)\cdot\sqrt[\alpha+\beta+\gamma+\delta]{a^{4\alpha+\gamma+\delta}b^{4\beta+\delta+\alpha}c^{4\gamma+\alpha+\beta}d^{4\delta+\beta+\gamma}}$$
by weighted AM-GM.
Choose $(\alpha,\beta,\gamma,\delta)=(9,7,1,3)$ (I got this by solving a linear equation system, feel free to ask me to elaborate; I am a bit in a rush at the moment) to get
$$9a^4 bc+7 b^4 c d+c^4da+3d^4 a b\geq20 a^2b^2cd.$$
Permuting $(\alpha,\beta,\gamma,\delta)$ cyclically around $(9,7,1,3)$ gives you analogous inequalities. Adding them up yields $$20 a^4bc+20 b^4cd+20 c^4da+20 d^4ab\geq 20a^2b^2cd+20b^2c^2da+20c^2d^2ab+20d^2a^2bc,$$ which is exactly what you want.