Let $\frac{x_{i+1}x_{i+2}}{x_i^2}=a_i$ Then we have to show that $\sum_{i=0}^n \frac{1}{1+a_i} \le n-1$ where the product of all $a_i$ is $1$. Lemma: $\frac{1}{1+x}+\frac{1}{1+y} \le 1+\frac{1}{1+xy}$ Proof: $(1+xy)(2+x+y) \le (x+1)(y+1)(2+xy)$ $2+x+y+2xy+x^2y+xy^2 \le 2xy+2x+2y+2+x^2y^2+x^2y+xy^2+xy$ All terms on the left have a corresponding term on the right, so the inequality is true. Using this, $\frac{1}{1+a_1}+\frac{1}{1+a_2} \le 1+\frac{1}{1+a_1a_2}$ $\frac{1}{1+a_1a_2} +\frac{1}{1+a_3} \le 1+\frac{1}{1+a_1a_2a_3}$ ... $\frac{1}{1+a_1a_2...a_{n-2}}+\frac{1}{1+a_{n-1}}\le 1+\frac{1}{1+a_1a_2...a_{n-1}}$ Adding all of these inequalities up and applying it to the original summation, it suffices to show that $n-2+\frac{1}{1+a_1a_2...a_{n-1}}+\frac{1}{1+a_n} \le n-1$ $\frac{1}{1+\frac{1}{a_n}}+\frac{1}{1+a_n} \le 1$ $\frac{a_n}{1+a_n}+\frac{1}{1+a_n} \le 1$ $\frac{a_n+1}{a_n+1} \le 1$ $1 \le 1$ Which is true. BTW, this problems was proposed by Canada xD. Cheers, Rofler

amparvardi wrote: Let $x_1, x_2, \cdots , x_n$ be positive numbers. Prove that \[\frac{x_1^2}{x_1^2+x_2x_3} + \frac{x_2^2}{x_2^2+x_3x_4} + \cdots +\frac{x_{n-1}^2}{x_{n-1}^2+x_nx_1} +\frac{x_n^2}{x_1^n+x_1x_2} \leq n-1\] hey !! equality dosen't hold ennigmation

Last fractional equals $ \frac {x_n^2}{x_n^2+x_1x_2}$

Ovchinnikov Denis wrote: Last fractional equals $ \frac {x_n^2}{x_n^2+x_1x_2}$ Yeah, thanks. Edited.

Is constant n-1 unimprovable?

amparvardi wrote: Let $x_1, x_2, \cdots , x_n$ be positive numbers. Prove that \[\frac{x_1^2}{x_1^2+x_2x_3} + \frac{x_2^2}{x_2^2+x_3x_4} + \cdots +\frac{x_{n-1}^2}{x_{n-1}^2+x_nx_1} +\frac{x_n^2}{x_n^2+x_1x_2} \leq n-1\] $ (*) \Leftrightarrow \sum {\left( 1 - \frac{x_1^2}{x_1^2+x_2x_3} \right)} \ge 1 $ ${ \Leftrightarrow \sum{\frac{x_2x_3}{x_1^2+x_2x_3}}} \ge 1 $ $ \Leftrightarrow \sum{\frac{1}{1+\frac{x_1^2}{x_2x_3}}} \ge 1 $ Let $ a_i = \frac{x_i^2}{x_{i+1}x_{i+2}} $ then $ \prod{a_i} = 1$ We have to prove that : $ \sum{\frac{1}{a_i+1}} \ge 1 $ There are $ a_i$ and $ a_j$ : $ a_ia_j \le 1 $. Then : $ \sum{\frac{1}{1+a_i}} \ge \frac{1}{a_i} + \frac{1}{a_j} = \frac{a_i+a_j+2}{a_ia_j+a_i+a_j+1} \ge 1 $ Hence we have done.

In fact holds next inequality: $ \sum_{i=0}^{n}\frac{1}{1+a_{i}}< n-1$ for $a_i>0; \prod_{i=1}^n=1$ Proving is like $Rofler$'s prove, but in inequality Quote: $ \frac{1}{1+x}+\frac{1}{1+y}\le 1+\frac{1}{1+xy} $ equality doesn't hold for positive $x, y$ Now proved that constant $n-1$ in right is unimprovable. Let $a_i= x$ for all $i \neq 1$ and $a_1=\frac{1}{x^n}$ Then $f(x)= \sum_{i=1}^{n} \frac{1}{1+a_i}= (n-1)\frac{x}{x+1}+\frac{1}{x^n+1}$ Let $ x \to \infty $ then $ \lim_{x \to \infty}f(x)=(n-1)\lim_{x \to \infty}\frac{x}{x+1} + \lim_{x \to \infty}\frac{1}{x^n+1}=n-1$ So $f(x) \to n-1 $when $x \to \infty$, so constant $n-1$ unimprovable, QED

Generalization : Let $x_1, x_2, \cdots , x_n$ be positive numbers ,$\alpha,\beta$ be real numbers. Prove that\[\frac{x_1^{\alpha +\beta}}{x_1^{\alpha +\beta}+x_2^{\alpha}x_3^{\beta}} + \frac{x_2^{\alpha +\beta}}{x_2^{\alpha +\beta}+x_3^{\alpha}x_4^{\beta}} + \cdots +\frac{x_{n-1}^{\alpha +\beta}}{x_{n-1}^{\alpha +\beta}+x_n^{\alpha}x_1^{\beta}} +\frac{x_n^{\alpha +\beta}}{x_n^{\alpha +\beta}+x_1^{\alpha}x_2^{\beta}} \leq n-1.\] http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&p=2861498

arqady wrote: Let $\frac{x_2x_3}{x_1^2}=\frac{a_2}{a_1}$, $\frac{x_3x_4}{x_2^2}=\frac{a_3}{a_2}$,..., $\frac{x_1x_2}{x_n^2}=\frac{a_1}{a_n}$, where $a_1$,$a_2$,..., $a_n$ are positives. Since for positives $a$, $b$ and $c$ holds $\frac{a}{a+b}\leq\frac{a+c}{a+b+c}$, we obtain: $\frac{x_{1}^{2}}{x_{1}^{2}+x_{2}x_{3}}+\frac{x_{2}^{2}}{x_{2}^{2}+x_{3}x_{4}}+.....+\frac{x_{n}^{2}}{x_{n}^{2}+x_{1}x_{2}}=$ $=\frac{a_1}{a_1+a_2}+\frac{a_2}{a_2+a_3}+...+\frac{a_n}{a_n+a_1}\leq$ $\leq\frac{a_1+a_3+...+a_n}{a_1+a_2+...+a_n}+\frac{a_1+a_2+a_4...+a_n}{a_1+a_2+...+a_n}+...+\frac{a_2+a_3+...+a_n}{a_1+a_2+...+a_n}=n-1$. http://www.artofproblemsolving.com/Forum/viewtopic.php?f=51&t=551349&p=3200657#p3200657