This is just AM-GM. \[ S_k=\sum_{i_1<...<i_k}a_{i_1}\cdots a_{i_k}\ge{n\choose k}\sqrt[{n\choose k}]{\prod_{i_1<...<i_k}a_{i_1}\cdots a_{i_k}}. \][I think Latex reaches its limits here ] The RHS of this has degree $k$ and is symmetrical in the $a_i$, so it must be $={n\choose k}(a_1\cdots a_n)^\frac kn$. Likewise $S_{n-k}\ge{n\choose n-k}(a_1\cdots a_n)^\frac{n-k}n$. The result follows.

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By Cauchy-Schwarz inequality \[ S_{k} S_{n-k}=\sum_{i_1<...<i_k} a_{i_1}\cdots a_{i_k} \sum_{i_1<...<i_k}\frac{S_n}{a_{i_1}\cdots a_{i_k}}\ge({n\choose k}\sqrt{S_{n}})^2={n\choose k}^2 S_{n} \]

Note that their are $n\choose k$ terms in $S_k$ and ${n\choose n-k}$ in $S_{n-k}$. By AM-GM we see that $$S_k=\sum_{i_1<\ldots<i_k}{a_{i_1}\cdot\ldots\cdot a_{i_k}}\ge {n\choose k}{\sqrt[n\choose k]{\prod_{i_1<\ldots<i_k}{a_{i_1}\cdot\ldots\cdot a_{i_k}}}}$$Now, fix any $a_i,i=1,\ldots,k$. There are now ${n-1\choose k-1}$ ways to combine $a_i$ with another $k-1$ factors, so $a_i$ appears ${n-1\choose k-1}$ terms in $S_k$ ant thus appears with power $n-1\choose k-1$ in the product on the right-hand-side. Therefore $${\sqrt[n\choose k]{\prod_{i_1<\ldots<i_k}{a_{i_1}\cdot\ldots\cdot a_{i_k}}}}=\left(a_1^{n-1\choose k-1}\cdot\ldots\cdot a_k^{n-1\choose k-1}\right)^\frac{1}{{n\choose k}}$$Now since $$\frac{{n-1\choose k-1}}{{n\choose k}}=\frac{k}{n}$$we get $$S_k\ge {n\choose k}\left(a_1\cdot\ldots\cdot a_k\right)^{\frac{k}{n}}$$In a similar fashion, we get $$S_{k-n}\ge \left(a_1\cdot\ldots\cdot a_k\right)^{\frac{n-k}{n}}$$and thus $$S_k\cdot S_{n-k}\ge {n\choose k}^2\cdot a_1 a_2 \cdots a_n$$

$\color{red}\boxed{\textbf{SOLUTION}}$ By $\textbf{AM-GM Inequality,}$ $$S_k=\sum_{1 \le i_1<\ldots<i_k \le n}{a_{i_1}\cdot\ldots\cdot a_{i_k}}\ge {n\choose k}{\sqrt[n\choose k]{\prod_{1 \le i_1<\ldots<i_k \le n}{a_{i_1}\cdot\ldots\cdot a_{i_k}}}}$$ And, $$S_{n-k}=\sum_{1\le i_{k+1}<\ldots<i_{k+j}\le n}{a_{i_{k+1}}\cdot\ldots\cdot a_{i_{k+j}}}\ge {n\choose k}{\sqrt[n\choose k]{\prod_{i_{k+1}<\ldots<i_{k+j}}{a_{i_{k+1}}\cdot\ldots\cdot a_{i_{k+j}}}}}$$ $\textbf{Note :}$ $\prod_{x \in (1,2...k,k+1,...k+j)}{a_{i_x}}=\prod_{1\le i \le n}{a_i}$ So, Multiplying Gives, $$S_k\cdot S_{n-k}\ge {n\choose k}^2\cdot a_1 a_2 \cdots a_n\blacksquare$$

True by Maclaurin's inequality.

Let $P$ be the product of all $a_i$. AM-GM gives \[S_k \ge \binom{n}{k} \sqrt[\binom{n}{k}]{P^{\binom{n-1}{k-1}}}\] and \[S_{n-k} \ge \binom{n}{n-k} \sqrt[\binom{n}{n-k}]{P^{\binom{n-1}{n-k-1}}}.\] Since \[\binom{n-1}{k-1} + \binom{n-1}{n-k-1} = \binom{n-1}{k-1} + \binom{n-1}{k} = \binom{n}{k},\] multiplying finishes. $\square$