Let $n$ be a fixed number, such that $n \geq 2$, and $n \in \mathbb{N}$. Let $\sigma_{i}(x_{1}, x_{2}, ..., x_{n})$ denote the i-th elementary symmetric polynomial in $x_{1}, x_{2}, ..., x_{n}$. Prove that the following inequality is satisfied,
$ (i + 1) (n - i) \sigma_{i}^{2} \geq (i +2) (n-i+1) \sigma_{i+1} \cdot \sigma_{i-1}$
I am trying to use Newton's inequality, but I am getting stuck
Newton's inequality is , $i (n - i) \sigma_{i}^{2} \geq (i +1) (n-i+1) \sigma_{i+1} \cdot \sigma_{i-1} $.
If I substitute for n, and i they seem to work out. For example, $n =3, i =1$, From Newton's inequality, we know $ 2 \; \sigma_{1}^{2} \geq 6 \; \sigma_{2} \cdot \sigma_{0}$ is true, and the inequality I want to prove evaluates to $ 4 \; \sigma_{1}^{2} \geq 9 \;\sigma_{2} \cdot \sigma_{0} $ which is true since $ 4 \; \sigma_{1}^{2} \geq 12 \; \sigma_{2} \cdot \sigma_{0}$ from Newton's inequality.
For positive variables we have $$\frac{\sigma_i^2}{\sigma_{i-1}\sigma_{i+1}}\geq\frac{(n-i+1)(i+1)}{(n-i)i}.$$ Thus, it's enough to prove that: $$\frac{(n-i+1)(i+1)}{(n-i)i}\geq\frac{(n-i+1)(i+2)}{(i+1)(n-i)}$$ or $$(i+1)^2\geq(i+2)i.$$ Can you end it now?