An inequality

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  • #28198

    swastik pramanik
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    Let \(a_1, a_2, \cdots, a_n\) be positive real numbers, and let \(S_k\) be the sum of products of \(a_1, a_2, \cdots, a_n\) taken \(k\) at a time. Show that $$S_kS_{n-k}\geq {n\choose k}^2a_1a_2\cdots a_n,$$ for \(k=1, 2,\cdots, n-1\).

    #28260

    Nitin Prasad
    Participant

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

    • This reply was modified 3 months, 3 weeks ago by  Nitin Prasad.
    • This reply was modified 3 months, 3 weeks ago by  Nitin Prasad.
    • This reply was modified 3 months, 3 weeks ago by  Nitin Prasad.
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