$\sum^n_{i=1} \frac{1+x_i}{1+x_{i+1}}=n+\sum^n_{i=1} \frac{x_i-x_{i+1}}{1+x_{i+1}}= n+\sum^n_{i=1} \frac{(x_i-x_{i+1})(1+x_{i+1})}{(1+x_{i+1})^2}\le n+\frac{1}{(1+a)^2}\cdot \sum^n_{i=1} (x_i-x_{i+1})(1+x_{i+1}),$ with equality when all $x_i$ are equal. We will now show that $\sum^n_{i=1} (x_i-x_{i+1})(1+x_{i+1})\le \sum^n_{i=1} (x_i-a)^2 \Longleftrightarrow \sum^n_{i=1} x_i x_{i+1}-\sum^n_{i=1} x_i^2\le \sum^n_{i=1}x_i^2+na^2-2a\sum^n_{i=1}x_i \Longleftrightarrow 2\sum^n_{i=1} x^2_i +na^2\ge \sum^n_{i=1} x_ix_{i+1} +2a\sum^n_{i=1} x_i \Longleftrightarrow \sum^n_{i=1} (x_i^2+x_{i+1}^2+a^2)\ge \sum^n_{i=1} [x_ix_{i+1}+a(x_i+x_{i+1})] \Longleftrightarrow \sum^n_{i=1} [a^2-a(x_i+x_{i+1})+x^2_i +x^2_{i+1}-x_i x_{i+1}]\ge 0.$ The latter is obvious because we have a quadratic in $a$ whose discriminant is $\le 0.$ Equality occurs when all $x_i$ are equal.

tk1 wrote: $ n+\sum^n_{i=1} \frac{(x_i-x_{i+1})(1+x_{i+1})}{(1+x_{i+1})^2}\le n+\frac{1}{(1+a)^2}\cdot \sum^n_{i=1} (x_i-x_{i+1})(1+x_{i+1})$ Why it's true?

\[ \sum^{n}_{i=1}\frac{1+x_i}{1+x_{i+1}}\le n+\frac{1}{(1+a)^2}\sum^{n}_{i=1}(x_i-a)^2\quad\quad (x_{n+1}=x_1), \] $let$ $\varepsilon _{i} = x_{i}-a$ then the inequality to prove is : $ \sum^{n}_{i=1}\frac{1+a+\varepsilon _i+\varepsilon _{i+1}-\varepsilon _{i+1}}{1+a+\varepsilon _{i+1}}\le n+\frac{1}{(1+a)^2}\sum^{n}_{i=1}(\varepsilon _i)^2\quad\quad $ $ \sum^{n}_{i=1}(\varepsilon _i)^2 \geq (1+a)^{2}\sum^{n}_{i=1}\frac{\varepsilon _i-\varepsilon _{i+1}}{1+a+\varepsilon _{i+1}}$ $set$ $m=a+1$ $then$ $ \sum^{n}_{i=1}(\varepsilon _i)^2 \geq \sum^{n}_{i=1}\frac{m^{2}\varepsilon _i}{m+\varepsilon _{i+1}}-\sum^{n}_{i=1}\frac{m^{2}\varepsilon _i}{m+\varepsilon _{i}}$ $\sum^{n}_{i=1}(\frac{\varepsilon _i}{m})^2+\frac{\varepsilon _i-\varepsilon _{i+1}}{m+\varepsilon _i}\geq 0$ with $\varepsilon _{0} =\varepsilon _{n}$ $ let$ $y_{i}=\frac{\varepsilon _{i}}{m} $ $then$ $\sum^{n}_{i=1}(y_{i}^2+\frac{y_i-y_{i+1}}{1+y_i})\geq 0$ $\sum^{n}_{i=1}(y_{i}+\frac{y_{i}^{3}}{1+y_{i}})\sum^{n}_{i=1}(\frac{y_{i-1}}{1+y_{i}}) $ which is true since $ \sum^{n}_{i=1}y_{i}\geq \sum^{n}_{i=1}(\frac{y_{i-1}}{1+y_{i}}) $ equality holds then when $y_{i}=0$ ie x_{i}=x_{i+1}

Let be $y_i=x_i-a\ge 0,i=1,2,\ldots n$, so the inequality can be rewrite as $\sum\limits_{cyc}\frac{y_{i+1}-y_i}{1+a+y_{i+1}}+\frac{1}{(1+a)^2}\sum\limits_{cyc}y_i^2\ge 0\Longleftrightarrow$ $\sum\limits_{cyc}\left(\frac{y_i}{1+a+y_i}-\frac{y_i}{1+a+y_{i+1}}\right)+\frac{1}{(1+a)^2}\sum\limits_{cyc}y_i^2\ge 0\Longleftrightarrow$ $\sum\limits_{cyc}\frac{y_iy_{i+1}}{(1+a+y_i)(1+a+y_{i+1})}+\sum\limits_{cyc}\left(\frac{1}{(1+a)^2}-\frac{1}{(1+a+y_i)(1+a+y_{i+1})}\right)y_i^2$ $\ge 0 $

arqady wrote: tk1 wrote: $ n+\sum^n_{i=1} \frac{(x_i-x_{i+1})(1+x_{i+1})}{(1+x_{i+1})^2}\le n+\frac{1}{(1+a)^2}\cdot \sum^n_{i=1} (x_i-x_{i+1})(1+x_{i+1})$ Why it's true? It isn't Thanks for pointing it out.