Feels a bit weird idk abt the equality cases, can someone check? @below ah lol

MrOreoJuice wrote:

Why are you allowed/what's the motivation to homogenize with $\sum n_i^3 = 1$? And how do you get \[\sum \frac{n_i^2}{Sn_i - n_i^3} \ge \frac{(\sum n_i)^2}{S^2 - 1}\]

My problem! By Cauchy Schwarz inequality, we have \[ \left(\sum_{i=1}^{k}\frac{n_i}{S-n_i^2}\right)(n_1+n_2 \cdots +n_k) \geq \left(\sum_{i=1}^{k}\sqrt{\frac{n_i^2}{S-n_i^2}}\right)^2 \] By the AM-GM inequality, we have $$\sqrt{\frac{S-n_i^2}{n_i^2}} \leq \frac{1}{2}\left(\frac{S-n_i^2}{n_i^2}+1\right)=\frac{1}{2}\left(\frac{S}{n_i^2}\right)$$ By flipping the fraction, this implies that $$\sqrt{\frac{n_i^2}{S-n_i^2}} \geq \frac{2n_i^2}{S}$$ $$\sum_{i=1}^{k}\sqrt{\frac{n_i^2}{S-n_i^2}}\geq \frac{2\displaystyle{\sum_{i=1}^{k}n_i^2}}{S}=2$$ Hence, $$\sum_{i=1}^{k}\frac{n_i}{S-n_i^2} \geq \frac{\left(\displaystyle{\sum_{i=1}^{k}\sqrt{\frac{n_i^2}{S-n_i^2}}}\right)^2}{n_1+n_2 \cdots +n_k}\geq\frac{4}{n_1+n_2 \cdots +n_k}$$ Equality case holds when $(n_1,n_2,...n_{k-2})$ all tends towards $0$ and $n_{k-1}=n_k$ Q.E.D

Let $n_i=\sqrt{x_i}$ and assume $x_1+\dots+x_k=1$. By CS, $$\left(\frac{\sqrt{x_1}}{1-x_1}+\dots+\frac{\sqrt{x_k}}{1-x_k}\right)\left (\sqrt{x_1}+\dots+\sqrt{x_k}\right) \geq \left (\sqrt{\frac{x_1}{1-x_1}}+\dots+\sqrt{\frac{x_k}{1-x_k}} \right)^2.$$Hence we just need to prove $$\sqrt{\frac{x_1}{1-x_1}}+\dots+\sqrt{\frac{x_k}{1-x_k}} \geq 2 \qquad (1)$$Let $f(x)=\sqrt{\frac{x}{1-x}}$ then $f''(x)=0\iff x=1/4$, hence we can apply EV. Let's assume $x_1=x_2=\dots=x_{k-1}$. Then $x_k=1-(k-1)x_1$ and $(1)$ becomes $$(k-1)\sqrt{\frac{x_1}{1-x_1}}+\sqrt{\frac{1}{(k-1)x_1}-1}\geq 2$$and this isn't very hard to prove for $0<x_1\leq \frac{1}{k-1}$.