This is a proof I came up with after looking at some suggestions. Hopefully it is complete.
Let us proceed by induction. First consider $k$ sequences of two terms each, $\{a_{i}^1\}_{i=1}^2,\ \{a_{i}^2\}_{i=1}^2, \ldots, \{a_{i}^k\}_{i=1}^2$. Each sequence in the following discussion shall be positive non-increasing.
Suppose for the sake of contradiction that the maximum sum is produced by a permutation where the maximal elements are not together. Without loss of generality, suppose $a^1_1$ and $a^2_1$ are separate:
$$a_1^1\cdot a^2_2\cdot\prod_{i=3}^k a^i_{\sigma_{i}(1)} + a_2^1\cdot a^2_1\cdot\prod_{i=3}^k a^i_{\sigma_{i}(2)}$$
We can assume that $a^j_1 \neq a^j_2$ for all sequences, otherwise we simply factor the term out and continue. Let us partition each summand into two smaller products.
First, $b_1$, shall contain all terms in the first summand such that $a_{\sigma_{i}(1)} > a_{\sigma_{i}(2)}$ while $s_1$ shall contain all terms in the first summand such that $a_{\sigma_{i}(1)} < a_{\sigma_{i}(2)}$. Similarly, we partition the second summand into $b_2$ containing the terms $a_{\sigma_{i}(1)} < a_{\sigma_{i}(2)}$ and $s_2$ containing the terms $a_{\sigma_{i}(1)} > a_{\sigma_{i}(2)}$.
This gives the sum as
$$a_1^1 \cdot a_2^2 \cdot s_1 \cdot b_1 + a_2^1 \cdot a_1^2 \cdot s_2 \cdot b_2$$
by construction, we have
$$a_2^2\cdot s_1 < a^2_1\cdot b_2,\ \ a_2^1\cdot s_2 < a_1^1\cdot b_1$$
and so by the rearrangement inequality, we have
$$a_1^1\cdot a_1^2 \cdot b_1 \cdot b_2 + a_2^1 \cdot a_2^2 \cdot s_1 \cdot s_2 \ge a_1^1\cdot a^2_2\cdot\prod_{i=2}^k a^i_{\sigma_{i}(1)} + a_2^1\cdot a^2_1\cdot\prod_{i=2}^k a^i_{\sigma_{i}(2)}$$
with equality if and only if $a_1^j = a_2^j$ for all sequences except $a^1$ or $a^2$. The process can be repeated to bring together all maximal elements. This contradicts the fact that the maximum sum is produced when maximal elements are separate. This handles the two term case.
Now suppose the inequality holds for sequences of length $n$ and let us look at the inequality for sequences of length $n+1$. There are two cases to handle. If the maximal elements are all together, then we simply remove the term and apply the inductive hypothesis immediately. If the maximal elements are not together, we can use the procedure in the base case repeatedly to bring them together, where upon we remove term and apply the inductive hypothesis.
OP is curious about whether a generalization is possible in comment.
Instead of the original inequality in question, I'm going to show a generalized version of that where $b$ can depend of $i$.
For any $x_1, \ldots, x_N \in \mathbb{R}_{+}$, let
$G(x_1,x_2,\ldots,x_N) = \left(\prod\limits_{k=1}^N x_k\right)^{1/N}$ be their geometric mean.
We recall following properties of the geometric mean:
- As a function, $G(x_1,\ldots,x_N)$ is increasing in each individual argument $x_k$.
- If one split $x_1,x_2,\ldots,x_N$ into two groups, we have
$$G(x_1,x_2,\ldots,x_N)^N = G(x_1,\ldots,x_M)^M G(x_{M+1},\ldots,x_N)^{N-M}$$
- In particular, if $N = 2M$ is even, this leads to
$$G(x_1,x_2,\ldots,x_N) = G(G(x_1,\ldots,x_M),G(x_{M+1},\ldots,x_N))$$
For any $n \ge 2$, let $S_n$ be the statement
For any $(a_1,\ldots,a_n), (b_1,\ldots,b_n) \in \mathbb{R}_{+}^n$,
$$
G(\{a_i + b_i\})
= \left(\prod_{k=1}^n (a_k+b_k)\right)^{1/n}
\ge G(\{a_i\}) + G(\{b_i\})
= \left(\prod_{k=1}^n a_k \right)^{1/n} + \left(\prod_{k=1}^n b_k \right)^{1/n}$$
$S_2$ is true.
Apply Cauchy Schwarz to $(\sqrt{a_1},\sqrt{b_1}), (\sqrt{a_2},\sqrt{b_2})$, we get
$$
\sqrt{(a_1+b_1)(a_2+b_2)}
= \sqrt{\left(\sqrt{a_1}^2+\sqrt{b_1}^2\right)\left(\sqrt{a_2}^2+\sqrt{b_2}^2\right)}
\ge \sqrt{a_1a_2} + \sqrt{b_1b_2}
$$
This is precisely $S_2$.
$S_2 \land S_n \implies S_{2n}$.
For any
$\begin{align}
(a_1,\ldots,a_{2n}) &= (a'_1,\ldots,a'_n,a''_1,\ldots,a''_n),\\
(b_1,\ldots,b_{2n}) &= (b'_1,\ldots,b'_n,b''_1,\ldots,b''_n)
\end{align}
\in \mathbb{R}_{+}^{2n}
$, we have
$$\begin{array}{rll}
G(\{ a_i + b_i \})
&= G(G(\{ a'_i + b'_i \}),G(\{a''_i + b''_i\})) & \color{blue}{\text{prop 3.}}\\
&\ge G(G(\{a'_i\}) + G(\{b'_i\}),G(\{a''_i\})+G(\{b''_i\})) & \color{blue}{S_n \text{ and prop 1.}}\\
&\ge G(G(\{a'_i\})G(\{a''_i\})) + G(G(\{b'_i\})G(\{b''_i\})) & \color{blue}{S_2}\\
&= G(\{a_i\}) + G(\{b_i\}) & \color{blue}{\text{prop 3.}}\\
\end{array}
$$
By principle of induction, $S_n$ is true whenever $n = 2^k$ is a power of $2$.
For general $n > 2$ but not a power of two, let $k$ be the integer
such that $2^{k-1} < n < 2^k$.
Let $\bar{a} = G(a_1,\ldots,a_n)$ and $\bar{b} = G(b_1,\ldots,b_n)$. Consider
following two $2^k$-tuples:
$$\begin{align}
( \tilde{a}_1,\ldots, \tilde{a}_{2^k})
&= ( a_1, a_2, \ldots, a_{n}, \bar{a}, \ldots, \bar{a} ),\\
( \tilde{b}_1,\ldots, \tilde{b}_{2^k})
&= ( b_1, b_2, \ldots, a_{n}, \bar{b}, \ldots, \bar{b} )
\end{align}
$$
It is easy to see
$G(\{\tilde{a}_i\}) = \bar{a}$ and
$G(\{\tilde{b}_i\}) = \bar{b}$.
Apply $S_{2^k}$ to the two $2^k$-tuples and raise both sides of result to $2^k$ power, we find
$$\begin{array}{rll} & G(\{ \tilde{a}_i + \tilde{b}_i \})^{2^k}
\ge (G(\{\tilde{a}_i\}) + G(\{\tilde{b}_i\}))^{2^k}\\
\iff & G(\{a_i + b_i\})^n (\bar{a}+\bar{b})^{2^k - n} \ge (\bar{a}+\bar{b})^{2^k}
& \color{blue}{\text{prop. 2 }}\\
\iff & G(\{a_i+b_i\}) \ge \bar{a} + \bar{b} = G(\{a_i\}) + G(\{b_i\})
\end{array}
$$
This implies $S_n$ is true for $n$ other than a power of $2$ too.
As a result, $S_n$ is true for all $n \ge 2$.
The inequality in question is a special case of this where all $b_i = b$.
Best Answer
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.