If $\frac{x}{y}+\frac{y}{x}\ge2$ and $\sum_{cyc}\frac{x}{y+z}\ge\frac32$, can we say $\sum_{cyc}\frac{w}{x+y+z}\ge\frac43$? We know that $$\frac{x}{y}+\frac{y}{x}\ge2$$ and $$\frac{x}{y+z}+\frac{y}{x+z}+\frac{z}{y+x}\ge\frac32$$ Can we say that
$$\frac{w}{x+y+z}+\frac{x}{w+y+z}+\frac{y}{x+w+z}+\frac{z}{x+y+w}\ge\frac43$$ Or in general, where there are $n$ variables can we say that their sum (in the same format as above) is no less than $\frac{n}{n-1}$$?$
All the variables used here are positive real numbers.
My thoughts:
I think that we might need induction here, and then we can try completing the squares to find the minimum value. Any help is greatly appreciated.

Problem: Let $x_i \ge 0, i= 1, 2, \cdots, n$. Prove that
$$\sum_{i=1}^n \frac{x_i}{S - x_i} \ge \frac{n}{n - 1}$$
where $S = \sum_{j=1}^n x_j$.
Proof.
Since the desired inequality is homogeneous, assume that $\sum_{j=1}^n x_j = 1$. We need to prove that
$$\sum_{i=1}^n \frac{x_i}{1 - x_i} \ge \frac{n}{n - 1}.$$
We have
$$\frac{x_i}{1 - x_i}
- \frac{1}{n - 1} - \frac{n^2}{(n-1)^2}\left(x_i - \frac{1}{n}\right) = \frac{(1-nx_i)^2}{(n-1)^2(1 - x_i)} \ge 0$$
which results in
$$\sum_{i=1}^n \frac{x_i}{1 - x_i} - \frac{n}{n-1} \ge \frac{n^2}{(n-1)^2}\sum_{i=1}^n\left(x_i - \frac{1}{n}\right) = 0.$$
We are done.
