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}$$?$
All the variables used here are positive real numbers.
My thoughts:
I think that we might need induction here, and then we can try completing the squares to find the minimum value. Any help is greatly appreciated.
 A: Proposition: $x_1,x_2,\dots, x_n\ge0$, then
$$\sum_{i=1}^{n}{\frac{x_i}{a-x_i}} \ge \frac{n}{n-1},$$
where $a=\sum_{i=1}^n x_i.$
Proof: Without loss of generality, we suppose $a=n$(otherwise we replace $x_i$ with $nx_i/a$, and we often do this in homogeneous inequalities), then it suffice to prove
$$\begin{aligned}
&\sum_{i=1}^{n}{\frac{x_i}{n-x_i}} \ge \frac{n}{n-1}\\
\iff &\sum_{i=1}^{n}{\Big(\frac{n}{n-x_i}-1\Big)} \ge \frac{n}{n-1} \\
\iff &\sum_{i=1}^{n}{\frac{n}{n-x_i}} \ge \frac{n}{n-1}+n=\frac{n^2}{n-1}\\
\iff &\sum_{i=1}^{n}{\frac{1}{n-x_i}} \ge \frac{n}{n-1}\\
\end{aligned}$$
By Cauchy-Schwarz Inequality, we have
$$\Big(\sum_{i=1}^{n}{\frac{1}{n-x_i}}\Big)\Big(\sum_{i=1}^n(n-x_i)\Big)\ge n^2,$$
where
$$\sum_{i=1}^n(n-x_i)=n^2-n, $$
Therefore
$$\sum_{i=1}^{n}{\frac{1}{n-x_i}}\ge\frac{n^2}{n^2-n}=\frac{n}{n-1}.$$
Now we are done with the proof.
A: Applying Cauchy-Schwars inequality, we have
$$\left(\overbrace{\frac{x}{y+z}+\frac{y}{x+z}+\frac{z}{y+x}}^A\right)\left(x(y+z)+y(x+z)+z(y+x)\right)\ge(x+y+z)^2\implies A≥\frac{(x+y+z)^2}{2(xy+yz+xz)}$$
Then let's apply AM-GM inequality:
$$
\begin{cases}x^2+y^2\ge 2xy\\x^2+z^2\ge 2xz\\ y^2+z^2\ge 2yz\end{cases}
$$
\begin{align}&2(x^2+y^2+z^2)\ge2(xy+yz+xz)\\
\implies &x^2+y^2+z^2 \ge xy+yz+xz\\
\implies &(x+y+z)^2\ge 3(xy+yz+xz)\end{align}
Hence, we obtain:
$$A≥\frac{(x+y+z)^2}{2(xy+yz+xz)}≥\frac{3(xy+yz+xz)}{2(xy+yz+xz)}=\frac 32$$

Now, let's apply Cauchy-Schwarz again:
$$\left(\overbrace{\frac{w}{x+y+z}+\frac{x}{w+y+z}+\frac{y}{x+w+z}+\frac{z}{x+y+w}}^A\right)\left(w(x+y+z)+x(w+y+z)+y(x+w+z)+z(x+y+w)\right)\implies A\ge \frac{(x+y+z+w)^2}{2(xy+xz+xw+yz+yw+zw)}$$
Let's apply AM-GM inequality again:
$$
\begin{cases}x^2+y^2\ge 2xy\\x^2+z^2\ge 2xz\\x^2+w^2\ge 2xw\\ y^2+z^2\ge2yz\\y^2+w^2\ge 2yw\\z^2+w^2\ge 2zw\end{cases}
$$
\begin{align}&3(x^2+y^2+z^2+w^2)\ge 2(xy+xz+xw+yz+yw+zw)\\
\implies &x^2+y^2+z^2+w^2 \ge \frac 23 (xy+xz+xw+yz+yw+zw)\\
\implies &(x+y+z+w)^2\ge \frac 83 (xy+xz+xw+yz+yw+zw)\end{align}
Finally, we have
$$A\ge \frac{(x+y+z+w)^2}{2(xy+xz+xw+yz+yw+zw)}\implies A\ge \frac 86=\frac 43$$
Now, can you generalize this method using a simple (just a few steps) combinatorics?
A: I'm uncertain if this proof is formally valid, but at least informally:
Consider a sequence of at least two positive real numbers $(a_1, a_2, \cdots a_k)$, and define $S_k = \sum a_i$. Then we claim:
$$A = \sum_{i=1}^{k} \frac{a_i}{S_k - a_i} \ge \frac{k}{k-1} \tag{1}$$
Note that $(1)$ implies that $A$ has its minimum when $a_1=a_2=\cdots=a_k$. In the simplest form, all $a_i=1$, $S_k=k$, and $A=k/(k-1)$.
Now, instead of $1$, WLOG set $a_1=1+z$, where $z$ is any real number $z>-1, z\ne 0$. Then, $S_k=k+z$. The restriction $z>-1$ prevents $a_1$ from becoming negative.
Let's find the value of $A$ with this changed $a_1$. It is:
$$A=\frac{z+1}{k-1} + \frac{k-1}{k+z-1}$$
The $k-1$ in the second fraction comes from $k-1$ copies of $1/(k-1+z)$. Both $k-1$ and $k+z-1$ must be positive (as $k \ge 2$ and $z>-1$).
Now, assume the contrary claim: that this set of $a_i$ gives a smaller total for $A$.
$$\frac{z+1}{k-1} + \frac{k-1}{k+z-1} < \frac{k}{k-1}$$
We can rearrange by
$$
\begin{align}
(z+1)(k+z-1) + (k-1)^2 &< k(k+z-1) \\
z^2 +kz+k^2-k &< k^2 +kz-k \\
z^2 &< 0
\end{align}
$$
Which is a pretty obvious contradiction. Hence, so long as adding $z$ doesn't make $a_1$ negative, $(1)$ holds.

However, this applies only to a single change from the set of $1$s. I'm not quite certain where to go from here, but I see other (possibly more complete) answers have been posted. Hopefully this partial proof is useful.
A: This answer is a generalization of the method used in the first answer given above.

Problem:
Prove that:
$$\sum_{1\le j\le n}{\frac{x_j}{\sum_{1\le i\le n} x_i-x_j}}\ge\frac{n}{n-1}$$

Proof:
Using AM-GM inequality we have,
$$\begin{align}&\sum_{1\le i<j\le n} (x_i^2+x_j^2)=(n-1)\sum_{1\le i\le n} x_i^2\ge 2\left(\sum_{1\le i < j \le n}x_ix_j\right)\\
\implies &\sum_{1\le i\le n} x_i^2\ge \frac 2{n-1}\left(\sum_{1\le i < j \le n}x_ix_j\right)\\ 
\implies &\left(\sum_{1\le i\le n} x_i\right)^2\ge\left( 2+\frac 2{n-1}\right)\left(\sum_{1\le i < j \le n}x_ix_j\right)\\
&\qquad\qquad\quad =\frac{2n}{n-1}\left(\sum_{1\le i < j \le n}x_ix_j\right)\end{align}$$
Finally, applying Cauchy-Schwars, we obtain
$$\begin{align}\left(\sum_{1\le j\le n}{\frac{x_j}{\sum_{1\le i\le n} x_i-x_j}}\right)\left(\sum_{1\le j\le n}^n \left({x_j}{\sum_{1\le i\le n} x_i-x_j}\right)\right)&\ge \frac{ \left(\sum_{1\le j\le n} x_i\right)^2}{2\left(\sum_{1\le i < j \le n}x_ix_j\right)}\\
&\ge \frac {\frac{2n}{n-1}\left(\sum_{1\le i < j \le n}x_ix_j\right)}{2\left(\sum_{1\le i < j \le n}x_ix_j\right)}\\
&=\boxed {\frac{n}{n-1}.}\end{align}$$
A: Here's a convex analysis approach. Let $x \in \mathbb{R}^d$ such that $x_i \ge 0$ and $\sum x_i > 0.$ We shall show (like the other answers) that $$ \sum_i \frac{x_i}{ \left(\sum_j x_j\right) -x_i} \ge \frac{n}{n-1}.$$

Without loss of generality (due to the homogeneity of $\frac{x_i}{\sum x_j - x_i}$ ), assume that $\sum x_i = 1$ . Then notice that $x$ is a vector in the simplex on $\mathbb{R}^d,$ i.e. it is a probability vector. We want to analyse the function
$$ f(x) := \sum_i \frac{x_i}{1 - x_i}.$$
Consider a random variable $Z$ which takes the value $x_i$ with probability $x_i$. Then $f(x) = \mathbb{E}[\frac{1}{1-Z}]$. Observe that $ u \mapsto \frac{1}{1-u}$ is convex over $[0,1]$. By Jensen's inequality, we conclude that $$ f(x) \ge \frac{1}{1 - \mathbb{E}[Z]} = \frac{1}{1 - \sum x_i^2}.$$ It is easy to show that $\min \sum x_i^2$ over the simplex in $\mathbb{R}^n$ is $1/n$. Immediately we conclude that $$ \min_x f(x) \ge \frac{1}{1 - 1/n} = \frac{n}{n-1}.$$ Of course, this is tight by taking all the $x_i$s to be equal.
A: 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.
