Throughout this part, $n$ denotes an integer greater than or equal to 3, and $p = [(n+1)/2]$ is the integer part of $(n+1)/2$. Let $a = (a_0, \ldots, a_{2n-2})$ be an element of $\mathbb{R}^{2n-1}$ and $M = H(a)$. We denote $\operatorname{Spo}(M) = (\lambda_1, \ldots, \lambda_n)$. We define two vectors $v = (v_1, \ldots, v_n)$ and $w = (w_1, \ldots, w_n)$ of $\mathbb{R}^n$ by $$\begin{cases} v_i = \sqrt{2i-1}\, a_{2(i-1)} \text{ and } w_i = \dfrac{1}{\sqrt{2i-1}} & \text{if } i \in \{1, \ldots, p\} \\ v_i = \sqrt{2n-2i+1}\, a_{2(i-1)} \text{ and } w_i = \dfrac{1}{\sqrt{2n-2i+1}} & \text{if } i \in \{p+1, \ldots, n\} \end{cases}$$ Finally, we set $K_n = n - \|w\|^2$. Show that $\displaystyle\sum_{1 \leqslant i < j \leqslant n} (\lambda_i - \lambda_j)^2 = n \sum_{i=1}^{n} \lambda_i^2 - \langle v, w \rangle^2$ and deduce the inequality: $$\sum_{1 \leqslant i < j \leqslant n} (\lambda_i - \lambda_j)^2 \geqslant K_n \sum_{i=1}^{n} \lambda_i^2 \tag{III.1}$$
Throughout this part, $n$ denotes an integer greater than or equal to 3, and $p = [(n+1)/2]$ is the integer part of $(n+1)/2$. Let $a = (a_0, \ldots, a_{2n-2})$ be an element of $\mathbb{R}^{2n-1}$ and $M = H(a)$. We denote $\operatorname{Spo}(M) = (\lambda_1, \ldots, \lambda_n)$.
We define two vectors $v = (v_1, \ldots, v_n)$ and $w = (w_1, \ldots, w_n)$ of $\mathbb{R}^n$ by
$$\begin{cases} v_i = \sqrt{2i-1}\, a_{2(i-1)} \text{ and } w_i = \dfrac{1}{\sqrt{2i-1}} & \text{if } i \in \{1, \ldots, p\} \\ v_i = \sqrt{2n-2i+1}\, a_{2(i-1)} \text{ and } w_i = \dfrac{1}{\sqrt{2n-2i+1}} & \text{if } i \in \{p+1, \ldots, n\} \end{cases}$$
Finally, we set $K_n = n - \|w\|^2$.
Show that $\displaystyle\sum_{1 \leqslant i < j \leqslant n} (\lambda_i - \lambda_j)^2 = n \sum_{i=1}^{n} \lambda_i^2 - \langle v, w \rangle^2$ and deduce the inequality:
$$\sum_{1 \leqslant i < j \leqslant n} (\lambda_i - \lambda_j)^2 \geqslant K_n \sum_{i=1}^{n} \lambda_i^2 \tag{III.1}$$