Let $f\in\mathbb{R}^n$ and $g\in\mathbb{R}^n$ be two orthogonal unit vectors such that $\sum_{i}{f_i}=\sum_{i}{g_i}=0$.

Question. Can we prove this? $$\frac{1}{2n-1} \le \frac{\sum_{i,j}\min((f_i-f_j)^2,(g_i-g_j)^2)}{\sum_{i,j}\max((f_i-f_j)^2,(g_i-g_j)^2)} $$

Let $f\in\mathbb{R}^n$ and $g\in\mathbb{R}^n$ be two orthogonal unit vectors such that $\sum_{i}{f_i}=\sum_{i}{g_i}=0$.

Question. Can we prove this? $$\frac{\sum_{\{i,j\}}\min((f_i-f_j)^2,(g_i-g_j)^2)}{\sum_{\{i,j\}}\max((f_i-f_j)^2,(g_i-g_j)^2)} \ge \frac{1}{2n-1}$$

Let $f\in\mathbb{R}^n$ and $g\in\mathbb{R}^n$ be two orthogonal unit vectors such that $\sum_{i}{f_i}=\sum_{i}{g_i}=0$. Can we prove that

$$\frac{1}{2n-1} \le \frac{\sum_{i,j}\min((f_i-f_j)^2,(g_i-g_j)^2)}{\sum_{i,j}\max((f_i-f_j)^2,(g_i-g_j)^2)} $$ ?

Let $f\in\mathbb{R}^n$ and $g\in\mathbb{R}^n$ be two orthogonal unit vectors such that $\sum_{i}{f_i}=\sum_{i}{g_i}=0$.

Question. Can we prove this? $$\frac{1}{2n-1} \le \frac{\sum_{i,j}\min((f_i-f_j)^2,(g_i-g_j)^2)}{\sum_{i,j}\max((f_i-f_j)^2,(g_i-g_j)^2)} $$