General case of :$\frac{a}{1+b}+\frac{b}{1+c}+\frac{c}{1+d}+\frac{d}{1+a}\leq \frac{a+b+c+d}{1+\frac{1}{4}(a+c)(b+d)}$ Here in my answer (Prove that $\frac{a}{1+b}+\frac{b}{1+c}+\frac{c}{1+d}+\frac{d}{1+a}\le2$ for $0 \le a, b, c, d \le 1$) I show the inequality :
Let $0\leq a,b,c,d\leq 1$:
$$\frac{a}{1+b}+\frac{b}{1+c}+\frac{c}{1+d}+\frac{d}{1+a}\leq \frac{a+b+c+d}{1+\frac{1}{4}(a+c)(b+d)}$$
Using buffalo's way
The Problem :
Let $0\leq x_i\leq 1$ be real such that $x_{n+1}=x_1$ and $n\geq 4$ prove or disprove that :
$$\sum_{i=1}^{n}\frac{x_{i}}{1+x_{i+1}}-\frac{\sum_{i=1}^{n}x_{i}}{1+\frac{\sum_{i=1}^{n}x_{i}x_{i+1}}{n}}\leq 0$$
As you can see we cannot use Buffalo's way in the general case because of prohibition of calculus
.
Perhaps we can use induction to show it .
Motivation :
It implies the general case (if true) of the inequality linked above .
My (funny) complicated way :
We have for $0< x,y\leq 1$ :
$$x\left(x+1\right)^{-y}\geq \frac{x}{xy+1}$$
So the LHS is :
$$\sum_{i=1}^{n}x_{i}\left(x_{i}+1\right)^{-\frac{x_{i+1}}{x_{i}}}$$
Now we use some constraint as $\frac{x_{i+1}}{x_i}=\frac{x_i+u}{k+x_i}$ and $1\leq i\leq n-1$ and $0<k$ and $0<u\leq 2k$  are constants next the function :
$$f(x)=x\left(x+1\right)^{-\frac{x+u}{k+x}}$$
Is concave on $(0,1]$ so using weighted Jensen's inequality the LHS is :
$$\left(\sum_{i=1}^{n-1}x_{i}\right)\left(\frac{\sum_{i=1}^{n-1}x_{i}^{2}}{\sum_{i=1}^{n-1}x_{i}}+1\right)^{-\frac{u\sum_{i=1}^{n-1}x_i+\sum_{i=1}^{n-1}x_{i}^{2}}{\left(k\sum_{i=1}^{n-1}x_{i}+\sum_{i=1}^{n-1}x_{i}^{2}\right)}}$$
Remains to compare with the RHS plus the last term and we need to  use some other constraint
Question :


How to (dis)prove it ?


Thanks in advance !

Note that, for all $x\in [0, 1]$,
$$1 - x/2 - \frac{1}{1 + x}
= \frac{x(1 - x)}{2(1 + x)}\ge 0.$$
We have
$$
 \sum_{\mathrm{cyc}}\frac{x_1}{1 + x_2}
 \le \sum_{\mathrm{cyc}} x_1(1 - x_2/2) 
 = \sum_{\mathrm{cyc}}x_1 - \frac12 \sum_{\mathrm{cyc}} x_1x_2.
$$
It suffices to prove that
\begin{align*}
 \frac{\sum_{\mathrm{cyc}}x_1}{1 + \frac{1}{n}\sum_{\mathrm{cyc}} x_1 x_2} \ge \sum_{\mathrm{cyc}}x_1 - \frac12 \sum_{\mathrm{cyc}} x_1x_2
\end{align*}
or
$$
\sum_{\mathrm{cyc}}x_1x_2 + n \ge 2\sum_{\mathrm{cyc}}x_1$$
or
$$\sum_{\mathrm{cyc}} (1 - x_1)(1 - x_2) \ge 0$$
which is true.
We are done.
