Define $$\sigma_k=\frac{S_k}{n\choose k}$$. Now we have following two classical result-
- Newton's inequality- $$\sigma_{k-1}\sigma_{k+1}\leq\sigma_{k}^2$$
- Maclaurian's inequality- $$(\sigma_{1})^{\frac{1}{1}}\geq(\sigma_{2})^{\frac{1}{2}}\geq\cdots\geq(\sigma_{n})^{\frac{1}{n}}$$
Now here we shall use second inequality to prove the given inequality which is equivalent to $$\sigma_{k}\sigma_{n-k}\geq \sigma_n$$. Now observe that-
$$(\sigma_{k})^{\frac{1}{k}}\geq (\sigma_{n})^{\frac{1}{n}}\Longleftrightarrow(\sigma_k)^n\geq(\sigma_n)^k$$
$$(\sigma_{n-k})^{\frac{1}{n-k}}\geq (\sigma_{n})^{\frac{1}{n}}\Longleftrightarrow(\sigma_{n-k})^n\geq(\sigma_n)^{n-k}$$
Hence $$(\sigma_k\sigma_{n-k})^n\geq \sigma_n^k\sigma_n^{n-k}=\sigma_n^n$$
i.e. $$\sigma_k\sigma_{n-k}\geq\sigma_n$$
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