Inequality of elementary symmetric polynomials

Solution 1:

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.

Solution 2:

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 $\lambda_1=\lambda_2=\lambda_3=\lambda_4.$