<s>[INST] 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}$$?$
All the variables used here are positive real numbers.
My thoughts:
I think that we might need induction here, and then we can try completing the squares to find the minimum value. Any help is greatly appreciated.
 [/INST] Applying Cauchy-Schwars inequality, we have
$$\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?
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