I don't understand the second proof of Muirhead’s inequality in page 10-11
First prove \eqref{constantMuirhead} using AM-GM:
Let $(c_i)_{i=1}^n$ be a sequence of real numbers such that $c_i \neq 0$ for some $i$ and $$ \sum_{i=1}^n c_i = 0. $$ For every pair of integers $(i,j)$ with $i,j \in \{1,\ldots, n\}$, there are $(n-1)!$ permutations $\sigma \in S_n$ with $\sigma(i) = j$ and so there are $(n-1)!$ permutations $\sigma$ such that $x_{\sigma(i)}^{c_i} = x_j^{c_i}$.
Therefore, \begin{align*} \prod_{\sigma \in S_n} x_{\sigma(1)}^{c_1} x_{\sigma(2)}^{c_2} \cdots x_{\sigma(n)}^{c_n} & = \prod_{\sigma \in S_n} \prod_{i=1}^n x_{\sigma(i)}^{c_i} = \prod_{i=1}^n \prod_{\sigma \in S_n} x_{\sigma(i)}^{c_i} \\ & = \prod_{i=1}^n \prod_{j =1}^n x_j^{(n-1)! c_i} = \prod_{j =1}^n \prod_{i =1}^n x_j^{(n-1)! c_i} \\ & = \prod_{j =1}^n x_j^{(n-1)! \sum_{i=1}^n c_i} = 1. \end{align*} Applying the arithmetic and geometric mean inequality to the nonconstant positive vector $\mathbf x$ gives $$ 1 = \left( \prod_{\sigma \in S_n} x_{\sigma(1)}^{c_1} x_{\sigma(2)}^{c_2} \cdots x_{\sigma(n)}^{c_n} \right)^{1/n!} < \frac{1}{n!}\sum_{\sigma \in S_n} x_{\sigma(1)}^{c_1} x_{\sigma(2)}^{c_2} \cdots x_{\sigma(n)}^{c_n} $$ and so $$ n! < \sum_{\sigma \in S_n} x_{\sigma(1)}^{c_1} x_{\sigma(2)}^{c_2} \cdots x_{\sigma(n)}^{c_n}. $$ Equivalently, $$\tag7\label{constantMuirhead} 0 < \sum_{\sigma \in S_n} \left( x_{\sigma(1)}^{c_1} x_{\sigma(2)}^{c_2} \cdots x_{\sigma(n)}^{c_n} - 1\right). $$
Then it proves Muirhead’s inequality using \eqref{constantMuirhead}:
Let $\mathbf a = \left( \begin{smallmatrix} a_1 \\ \vdots \\ a_n \end{smallmatrix}\right)$ and $\mathbf b = \left( \begin{smallmatrix} b_1 \\ \vdots \\ b_n \end{smallmatrix}\right)$.
For $i \in\{1,\ldots, n\}$, let $$ c_i = a_i-b_i. $$ The strict majorization relation $\mathbf b \prec \mathbf a$ implies that $c_i \neq 0$ for some $i$ and $$ \sum_{i=1}^n c_i = \sum_{i=1}^n (a_i-b_i) = \sum_{i=1}^n a_i - \sum_{i=1}^n b_i = 0. $$ We have \begin{align} \sum_{\sigma \in S_n} \prod_{i=1}^n x_{\sigma(i)}^{a_i} & = \sum_{\sigma \in S_n} \prod_{i=1}^n x_{\sigma(i)}^{b_i+c_i} = \sum_{\sigma \in S_n} \prod_{i=1}^n x_{\sigma(i)}^{b_i} \prod_{i=1}^n x_{\sigma(i)}^{c_i}. \end{align} Inequality \eqref{constantMuirhead} implies
\begin{align} [\mathbf x^{\mathbf a}]_{S_n} - [\mathbf x^{\mathbf b}]_{S_n} & = \sum_{\sigma \in S_n} \prod_{i=1}^n x_{\sigma(i)}^{a_i} - \sum_{\sigma \in S_n} \prod_{i=1}^n x_{\sigma(i)}^{b_i} \\ & = \sum_{\sigma \in S_n} \prod_{i=1}^n x_{\sigma(i)}^{b_i} \prod_{i=1}^n x_{\sigma(i)}^{c_i} - \sum_{\sigma \in S_n} \prod_{i=1}^n x_{\sigma(i)}^{b_i} \\ & = \sum_{\sigma \in S_n} \prod_{i=1}^n x_{\sigma(i)}^{b_i} \left( \prod_{i=1}^n x_{\sigma(i)}^{c_i} -1\right) \\ & > 0. \end{align} This completes the proof.
I can't prove the last step using \eqref{constantMuirhead} $$\sum_{\sigma \in S_n} \prod_{i=1}^n x_{\sigma(i)}^{b_i} \left( \prod_{i=1}^n x_{\sigma(i)}^{c_i} -1\right) > 0 $$ An equivalent form of \eqref{constantMuirhead} is$$\sum_{\sigma \in S_n} \left( \prod_{i=1}^n x_{\sigma(i)}^{c_i} -1\right) > 0 $$ It is not the case that $\left(\prod_{i=1}^n x_{\sigma(i)}^{c_i} -1\right)>0$ for all $\sigma \in S_n$.