Inequality proof by using the Muirhead inequality.
Let $\alpha_1\geq\alpha_2\geq...\geq\alpha_n\geq0$ and $\beta_1\geq\beta_2\geq...\geq\beta_n$.
We'll write $$(\alpha_1,\alpha_2,..., \alpha_n)\succ(\beta_1,\beta_2,...,\beta_n)$$ iff $$\alpha_1\geq\beta_1,$$ $$\alpha_1+\alpha_2\geq\beta_1+\beta_2,$$ $$.$$ $$.$$ $$.$$ $$\alpha_1+\alpha_2+...+\alpha_n=\beta_1+\beta_2+...+\beta_n.$$
- Muirhead inequality it's the following.
Let $x_1$, $x_2$,..., $x_n$ be positive numbers and $(\alpha_1,\alpha_2,..., \alpha_n)\succ(\beta_1,\beta_2,...,\beta_n)$. Prove that: $$\sum_{sym}\prod_{i=1}^nx_i^{\alpha_i}\geq\sum_{sym}\prod_{i=1}^nx_i^{\beta_i}$$
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