General case of :$\frac{a}{1+b}+\frac{b}{1+c}+\frac{c}{1+d}+\frac{d}{1+a}\leq \frac{a+b+c+d}{1+\frac{1}{4}(a+c)(b+d)}$

Question

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 !

Answer 1

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.