Wednesday, November 9th

Problem
Let $x_1, x_2, \dots, x_n$ be positive reals and let $S=x_1+x_2+...+x_n.$ Prove that 

$(1+x_1)(1+x_2)...(1+x_n)\leq 1+\frac{S}{1!}+\frac{S^2}{2!}+...+\frac{S^n}{n!}$.

Solution
By the McLaurin inequalities we have that

$S_1^{j}\geq S_j$

where $S_i$ is the $i$th elementary symmetric average. Thus 

$S^j\frac{\binom{n}{j}}{n^j}\geq S_j\binom{n}{j}$

and is it easily seen that 

$\frac{1}{j!}\geq \frac{\binom{n}{j}}{n^j}$

therefore

$\frac{S^j}{j!}\geq S_j\binom{n}{j}=\sigma_j$

where $\sigma_j$ is the $j$th elementary symmetric function. Thus expanding the LHS of the inequality gives the result.