$n\geq 2$ and $\sum_{i=1}^{n}x_i^2=1 \Rightarrow \sum_{i=1}^{n}\frac{|x_i|}{\sqrt{1-x_i^2}}\geq 2$

Question

The original inequality is expressed as $$\sum_{i=1}^{n}\frac{a_i}{\sum_{k=1}^{n}a_kb_k-a_ib_i}\geq\frac{4}{\sum_{i=1}^{n}b_i}$$ where $a_i, b_i>0$ for all $i\in\{1,2,\ldots,n\}$. To simplify and further explore the inequality, it's natural to introduce constraints $\sum b_k=1$ and $\sum a_kb_k=1$. Applying Cauchy's inequality yields: $$\sum b_i\sum\frac{1}{b_i}\frac{a_ib_i}{1-a_ib_i}\geq\left(\sum\sqrt\frac{a_ib_i}{1-a_ib_i}\right)^2$$ It remains to verify $$\sum\sqrt\frac{a_ib_i}{1-a_ib_i}\geq2$$ My guess is that$\sum\frac{|x_i|}{\sqrt{1-x_i^2}}\geq\sum\frac{\frac{1}{\sqrt{n}}}{\sqrt{1-\frac{1}{n}}}=\frac{n}{\sqrt{n-1}}\geq2$.

Answer 1

Now, by AM-GM $$\sum_{i=1}^n\frac{|x_i|}{\sqrt{1-x_i^2}}=\sum_{i=1}^n\frac{2x_i^2}{2\sqrt{x_i^2(1-x_i^2)}}\geq\sum_{i=1}^n\frac{2x_i^2}{x_i^2+1-x_i^2}=2.$$