Talk:Muirhead's Inequality

Rewriting the section about the proof

Note that in the argument that was in that section before, there was a mistake that is not easy to fix. In the step

\[\left(\sum_{sym}\prod_{i=1}^{n}x_i^{b_i}\right)\left(\sum_{sym}x_1^{c_1}x_2^{c_2}\dotsm x_n^{c_n}-1\right)=\sum_{sym}\prod_{i=1}^{n}x_i^{b_i+c_i}-\prod_{i=1}^{n}x_i^{b_i}\]

Distributing the parentheses on the left we get many other terms in which the exponent $b_i$ doesn't get added to the exponent $c_i$, but some other $c_{i'}$. The right hand side is what we would get if we only multiplied each term from the first parentheses, corresponding to a particular permutation of the variables, to the term in the second parenthesis that corresponds to the same permutation of the variable. The distributive property multiplies every term of the first parenthesis with every term of the second parenthesis.