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Let $\lambda=(\lambda_1,\lambda_2,\lambda_3,\lambda_4)$ with $\lambda_i>0$ for $i=1,2,3,4$. Let $$\sigma_k(\lambda)=\sum_{1\leq i_1<i_2\cdots<i_k\leq4}\lambda_{i_1}\lambda_{i_2}\cdots\lambda_{i_k},\quad k=1,2,3,4$$ be the elementary symmetric polynomials. Prove the following inequality $$\sigma_2^2-3\sigma_1\sigma_3+12\sigma_4\geq0.$$ Furthermore, characterize the equality case.

To deal with inequalities of symmetric polynomials, Newton Maclaurin type inequalities also be useful. I tried to use the following inequlities $$\sigma_1/4\geq\frac{\sigma_2/6}{\sigma_1/4}\geq\frac{\sigma_3/4}{\sigma_2/6}\geq\frac{\sigma_4}{\sigma_3/4},$$ but failed.

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  • $\begingroup$ What exactly is the positive cone of $\mathbb R^4$ ? Or of $\mathbb R^3,$ for that matter ? And is $\lambda=(\lambda_1,\lambda_2,\lambda_3,\lambda_4)$ ? Or are you asking us to add and multiply quaternions ? $\endgroup$
    – Lucian
    May 16, 2015 at 21:21
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    $\begingroup$ There is no $k$ in your definition of $\sigma_k$. $\endgroup$
    – Alex R.
    May 16, 2015 at 22:58

2 Answers 2

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Hint: This method works often with symmetric inequalities. Assume an order $$0 \le \lambda_1 \le \lambda_2 \le \lambda_3 \le \lambda_4$$ and write $$\lambda_k = l_1 +\ldots + l_k$$ where $l_i$ are all $\ge 0$. Substitute in your inequality and see what you get. You will also notice that the equality is achieved if an only if at least three of the $\lambda_i$'s are equal.

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For simplicity of presentation, let $$E_k(\lambda)=\binom{4}{k}^{-1}\sigma_k(\lambda),\quad 1\le k\le 4,$$ be the normalized elementary symmetric polynomials.

From Newton-Maclaurin inequalities, we get $$E_1 E_3\le E_2^2\le E_4.$$

Hence \begin{align*} \frac{1}{12}[\sigma_2^2-3\sigma_1\sigma_3+12\sigma_4] &=3E_2^2-4E_1 E_3+E_4 \\ &\ge 3E_2^2-4E_2^2+E_4=E_4-E_2^2\\ &\ge0, \end{align*} with the equality holds if and only if $\lambda_1=\lambda_2=\lambda_3=\lambda_4.$

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