The problem :
In the hint of problem 3.50 it is mentioned that k-th elementary function defined as follows $$S_k (x)=\sum_{1\leq i_1 <i_2\cdots < i_k\leq n} x_{i_1}x_{i_2}\cdots x_{i_n} $$ is concave. Since we know that logrithim of a positive concave function is concave therefore $S_k (x) $ is concave. But $S_2 (x)=xy $ and $\log (xy) $ is convex. Which point am I missing here? Any help will be much appreciated.
If you expand the right hand side in the link, you can show that \begin{equation} a_{k}(\lambda) = \sum_{1 \leq j_1 \leq \ldots \leq j_k \leq n} (-\lambda_{j_1})(-\lambda_{j_2}) \ldots (-\lambda_{j_k}) \qquad 1 \leq k \leq n \end{equation} or simply \begin{equation} a_{k}(\lambda) = (-1)^k \sum_{1 \leq j_1 \leq \ldots \leq j_k \leq n} \lambda_{j_1}\lambda_{j_2} \ldots \lambda_{j_k} \qquad 1 \leq k \leq n \end{equation} You can prove that $(a_{k}(\lambda))^{\frac{1}{k}}$ is concave on $-R_{+}^n$ because each of the $\lambda$'s are positive. Now since $(a_{k}(\lambda))^{\frac{1}{k}}$ is concave on $-R_{+}^n$, then its log is log-concave.