APMO

$$\left(\genfrac{}{}{0ex}{}{n}{k}\right)a_1{a}_2\dots a_n$$ $$=\sum _{1\le i_1<i_2<\dots <i_k\le n}{a}_{i_1}{a}_{i_2}\dots a_{i_k}\cdot a_1{a}_2\dots a_n/a_{i_1}{a}_{i_2}\dots a_{i_k}$$ (and using the Cauchy-Schwarz inequality) $$\begin{array}{rl}& \le {\left(\sum _{1\le i_1<i_2<\dots <i_k\le n}{a}_{i_1}{a}_{i_2}\dots a_{i_k}\right)}^{\frac{1}{2}}\cdot {\left(\sum _{1\le i_1<i_2<\dots <i_k\le n}{a}_1{a}_2\dots a_n/a_{i_1}{a}_{i_2}\dots a_{i_k}\right)}^{\frac{1}{2}}\\ & =S_k^{\frac{1}{2}}\cdot S_{n-k}^{\frac{1}{2}}.\end{array}$$ Therefore $${\left(\genfrac{}{}{0ex}{}{n}{k}\right)}^2{a}_1{a}_2\dots a_n\le S_k{S}_{n-k}$$ q.e.d.

Solution 2:

Write $S_k$ as ${\sum }_{i=1}^{\left(\genfrac{}{}{0ex}{}{n}{k}\right)}{t}_i$. Then $$\begin{array}{rl}& \begin{array}{rl}{S}_{n-k}& =\left(\prod _{m=1}^n{a}_m\right)\left(\sum _{i=1}^{\left(\genfrac{}{}{0ex}{}{n}{k}\right)}\frac{1}{t_i}\right)\\ \text{ so that }{S}_k{S}_{n-k}& =\left(\prod _{m=1}^n{a}_m\right)\left(\sum _{i=1}^{\left(\genfrac{}{}{0ex}{}{n}{k}\right)}{t}_i\right)\left(\sum _{j=1}^{\left(\genfrac{}{}{0ex}{}{n}{k}\right)}\frac{1}{t_j}\right)\\ & =\left(\prod _{m=1}^n{a}_m\right)\left[\sum _{i=1}^{\left(\genfrac{}{}{0ex}{}{n}{k}\right)}1+\sum _{i=1}^{\left(\genfrac{}{}{0ex}{}{n}{k}\right)}\sum _{j=1}^{\left(\genfrac{}{}{0ex}{}{n}{k}\right)}\frac{t_i}{t_j}\right]\\ & =\left(\prod _{m=1}^n{a}_m\right)\left[\left(\genfrac{}{}{0ex}{}{n}{k}\right)+\sum _{i,j}\left(\frac{t_i}{t_j}+\frac{t_j}{t_i}\right)\right]\end{array}.\end{array}$$ As there are $$\frac{{\left(\genfrac{}{}{0ex}{}{n}{k}\right)}^2-\left(\genfrac{}{}{0ex}{}{n}{k}\right)}{2}$$ terms in the sum $$\begin{array}{rl}{S}_k{S}_{n-k}& \ge \left(\prod _{m=1}^n{a}_m\right)\left[\left(\genfrac{}{}{0ex}{}{n}{k}\right)+2\cdot \frac{{\left(\genfrac{}{}{0ex}{}{n}{k}\right)}^2-\left(\genfrac{}{}{0ex}{}{n}{k}\right)}{2}\right]\\ & ={\left(\genfrac{}{}{0ex}{}{n}{k}\right)}^2\left(\prod _{m=1}^n{a}_m\right)\end{array}$$ since $\frac{t_i}{t_j}+\frac{t_j}{t_i}\ge 2$ for $t_i,t_j>0$.

There are $\left(\genfrac{}{}{0ex}{}{n}{k}\right)$ products of the $a_i$ taken $k$ at a time. Amongst these products any given $a_i$ will appear $\left(\genfrac{}{}{0ex}{}{n-1}{k-1}\right)$ times, since $\left(\genfrac{}{}{0ex}{}{n-1}{k-1}\right)$ is the number of ways of choosing the other factors of the product. So the AM/GM inequality gives $$\frac{S_k}{\left(\genfrac{}{}{0ex}{}{n}{k}\right)}\ge {\left[\prod _{i=1}^n{a}_i^{\left(\genfrac{}{}{0ex}{}{n-1}{k-1}\right)}\right]}^{\frac{1}{\left(\genfrac{}{}{0ex}{}{n}{k}\right)}}$$ But $\left(\genfrac{}{}{0ex}{}{n}{k}\right)=\frac{n}{k}\left(\genfrac{}{}{0ex}{}{n-1}{k-1}\right)$, leading to $$S_k\ge \left(\genfrac{}{}{0ex}{}{n}{k}\right){\left(\prod _{i=1}^n{a}_i\right)}^{\frac{k}{n}}$$ Hence $$S_k{S}_{n-k}\ge \left(\genfrac{}{}{0ex}{}{n}{k}\right){\left(\prod _{i=1}^n{a}_i\right)}^{\frac{k}{n}}\left(\genfrac{}{}{0ex}{}{n}{n-k}\right){\left(\prod _{i=1}^n{a}_i\right)}^{\frac{n-k}{n}}={\left(\genfrac{}{}{0ex}{}{n}{k}\right)}^2\left(\prod _i^n{a}_i\right).$$