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.
We prove the inequality by using the rearrangement inequality
We observe that
- the numerator is an increasing function. Indeed, let's denote
$$f(x) = x^5+4x^2+3x+8$$
then
$$f'(x) = 5x^4+8x+3 >0 \hspace{1cm} \text{for }x>0$$
- the denominator $x \mapsto(x+1)^2$ is an increasing function, then the function $x \mapsto \frac{1}{(x+1)^2}$ is decreasing.
Then, WOLG, we assume that $x \ge y \ge z$, by using the rearrangement inequality, we have
$$\text{LHS} =\sum_{x,y,z} \frac{x^5+4x^2+3x+8}{(\color{red}{y}+1)^2} \ge \sum_{x,y,z} \frac{x^5+4x^2+3x+8}{(\color{red}{x}+1)^2} \tag{1}$$
And it easy to prove that
$$\frac{x^5+4x^2+3x+8}{(x+1)^2} - 4 = \frac{(x-1)^2 (x^3+2x^2+3x+4)}{(x+1)^2} \ge0 \tag{2}$$
From $(1),(2)$, we deduce that $\text{LHS} \ge 12$ and the equality occurs if and only if $x=y=z=1$.
Best Answer
The cyclically symmetric inequality is equivalent to: $$\color{red}{\left(\frac{a^2}{b^2}+\frac{b^2}{a^2}-2 \right)} + \color{blue}{\left(\frac{b^2}{c^2}+\frac{c^2}{a^2}-\frac{b^2}{a^2}-1 \right)}+ \color{green}{\left(\frac{9abc}{4(a^3+b^3+c^3)}-\frac34\right)} \geqslant 0$$ $$\iff \color{red}{\frac{(a^2-b^2)^2}{a^2b^2}} + \color{blue}{\frac{(a^2-c^2)(b^2-c^2)}{a^2c^2}}+\color{green}{\frac{9abc-3(a^3+b^3+c^3)}{4(a^3+b^3+c^3)}}\geqslant 0$$ As $a^3+b^3+c^3-3abc=(a+b+c)((a-b)^2+(a-c)(b-c))$, we have above $\iff$ $$(a-b)^2\left(\color{red}{\frac{(a+b)^2}{a^2b^2}}-\color{green}{\frac{3(a+b+c)}{4(a^3+b^3+c^3)}} \right)+(a-c)(b-c)\left(\color{blue}{\frac{(a+c)(b+c)}{a^2c^2}} - \color{green}{\frac{3(a+b+c)}{4(a^3+b^3+c^3)}}\right)\geqslant 0$$
Now due to symmetry, we may assume $c=\min (a, b, c)$, hence it remains to show that under this condition, both $$\color{red}{\frac{(a+b)^2}{a^2b^2}}-\color{green}{\frac{3(a+b+c)}{4(a^3+b^3+c^3)}} \geqslant 0, \qquad \color{blue}{\frac{(a+c)(b+c)}{a^2c^2}} - \color{green}{\frac{3(a+b+c)}{4(a^3+b^3+c^3)}} \geqslant 0$$
However $(a^3+b^3)(a+b)\geqslant (a^2+b^2)^2\geqslant 4a^2b^2 \implies$ $$\color{red}{\frac{(a+b)^2}{a^2b^2}}\geqslant 4\frac{a+b}{a^3+b^3}> 4\frac{a+b}{a^3+b^3+c^3}\geqslant \frac83\frac{a+b+c}{a^3+b^3+c^3}> \color{green}{\frac{3(a+b+c)}{4(a^3+b^3+c^3)}}$$ And $3(a^3+b^3+c^3)\geqslant (a+b+c)(a^2+b^2+c^2) \implies$ $$4(a^3+b^3+c^3)(a+c)(b+c)\geqslant \frac43(a+b+c)(a+c)(b+c)(a^2+b^2+c^2) \geqslant \frac43(a+b+c)(2c)(2c)(a^2)\geqslant 3(a+b+c)a^2c^2$$ $$\implies \color{blue}{\frac{(a+c)(b+c)}{a^2c^2}}\geqslant \color{green}{\frac{3(a+b+c)}{4(a^3+b^3+c^3)}}$$ Hence the inequality holds true, with equality when $a=b=c$.