[Nitin Prasad]

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. $$\sigma_k\sigma_{n-k}\geq\sigma_n$$

[Author]

Posts