Titu + AM-GM

I am wrong

navi_09220114 wrote: Split the numerator then use T2 lemma Yes we have to use T2 lemma to both parts and then use the simple idea$ n+\sum_{i=2}^{n}x_ix_{i-1}<n $ .But then when does equality hold's.

navi_09220114 wrote: Split the numerator then use T2 lemma This gives $\frac{1+x_{1}^2}{1+x_{1}x_{2}}+\frac{1+x_{2}^2}{1+x_{2}x_{3}}+...+\frac{1+x_{n}^2}{1+x_{n}x_{1}}\ge \frac{n^2+(x_1+x_2+...+x_n)^2}{n+x_1x_2+x_2x_3+...+x_nx_1}$. What's your proof it's greater than $n$? Because I don't see this at all.

ARMO 2018 wrote: Let $n\geq 2$ and $x_{1},x_{2},....,x_{n}$ positive real numbers. Prove that $$\frac{1+x_{1}^2}{1+x_{1}x_{2}}+\frac{1+x_{2}^2}{1+x_{2}x_{3}}+\cdots+\frac{1+x_{n}^2}{1+x_{n}x_{1}}\geq n$$

Davrbek wrote: Let $n\geq 2$ and $x_{1},x_{2},....,x_{n}$ positive real numbers.Prove that $\frac{1+x_{1}^2}{1+x_{1}x_{2}}+\frac{1+x_{2}^2}{1+x_{2}x_{3}}+...+\frac{1+x_{n}^2}{1+x_{n}x_{1}}\geq n$ $$(1+ x^2_i)(1 + x^2_{i+1}) > (1 + x_ix_{i+1})^2$$$$\frac{1+x_{1}^2}{1+x_{1}x_{2}}+\frac{1+x_{2}^2}{1+x_{2}x_{3}}+...+\frac{1+x_{n}^2}{1+x_{n}x_{1}}\geq n\sqrt[n]{\frac{1+x_{1}^2}{1+x_{1}x_{2}}\cdot \frac{1+x_{2}^2}{1+x_{2}x_{3}}\cdots\frac{1+x_{n}^2}{1+x_{n}x_{1}}}\geq n$$Polish MO 2017/2018

vickyricky wrote: navi_09220114 wrote: Split the numerator then use T2 lemma Yes we have to use T2 lemma to both parts and then use the simple idea$ n+\sum_{i=2}^{n}x_ix_{i-1}<n $ .But then when does equality hold's. Easy to see your solution is incorrect. You can't reach equality using your simple idea. At the same time putting $x_1=x_2=...=x_n$ we get equality in the given inequality.

Poland MO 2018 First Round Task 10 Let $n\geq 3$ and $x_{1},x_{2},....,x_{n}$ positive real numbers.Prove that $$\frac{1+x_{1}^2}{x_{2}+x_{3}}+\frac{1+x_{2}^2}{x_{3}+x_{4}}+...+\frac{1+x_{n}^2}{x_{1}+x_{2}}\geq n$$I think this task inspired Russia MO

Davrbek wrote: Let $n\geq 2$ and $x_{1},x_{2},....,x_{n}$ positive real numbers.Prove that $\frac{1+x_{1}^2}{1+x_{1}x_{2}}+\frac{1+x_{2}^2}{1+x_{2}x_{3}}+...+\frac{1+x_{n}^2}{1+x_{n}x_{1}}\geq n$ Notice that $1+ab \le \sqrt{(1+a^2)(1+b^2)}$ for all $a,b$; hence \begin{align*} \frac{1+x_{1}^2}{1+x_{1}x_{2}}+\frac{1+x_{2}^2}{1+x_{2}x_{3}}+...+\frac{1+x_{n}^2}{1+x_{n}x_{1}} & \ge \sqrt{\frac{1+x_1^2}{1+x_2^2}}+\sqrt{\frac{1+x_2^2}{1+x_3^2}}+\dots+\sqrt{\frac{1+x_n^2}{1+x_1^2}} \\ & \ge n \end{align*}with the last inequality following from AM-GM. Equality occurs when $x_1=\dots=x_n$.

anantmudgal09 wrote: Davrbek wrote: Let $n\geq 2$ and $x_{1},x_{2},....,x_{n}$ positive real numbers.Prove that $\frac{1+x_{1}^2}{1+x_{1}x_{2}}+\frac{1+x_{2}^2}{1+x_{2}x_{3}}+...+\frac{1+x_{n}^2}{1+x_{n}x_{1}}\geq n$ Notice that $1+ab \le \sqrt{(1+a^2)(1+b^2)}$ for all $a,b$; hence \begin{align*} \frac{1+x_{1}^2}{1+x_{1}x_{2}}+\frac{1+x_{2}^2}{1+x_{2}x_{3}}+...+\frac{1+x_{n}^2}{1+x_{n}x_{1}} & \ge \sqrt{\frac{1+x_1^2}{1+x_2^2}}+\sqrt{\frac{1+x_2^2}{1+x_3^2}}+\dots+\sqrt{\frac{1+x_n^2}{1+x_1^2}} \\ & \ge n \end{align*}with the last inequality following from AM-GM. Equality occurs when $x_1=\dots=x_n$. Nice .

Also IMAR 2018 has a strong statement. Let$ P\in R[X]$ with all positive coefficients.Prove that\[\frac{1+P(x_{1})^2}{1+P(x_{1}t(x_{1}))}+\frac{1+P(x_{2})^2}{1+P(x_{2}t(x_{2}))}+\cdots+\frac{1+P(x_{n})^2}{1+P(x_{n}t(x_{n}))}\geq n.\],where $ t$ is a permutation of nubers $ x_1,x_2,..x_n$. The proof is identic with russian problem.

Davrbek wrote: Let $n\geq 2$ and $x_{1},x_{2},\ldots,x_{n}$ positive real numbers. Prove that \[\frac{1+x_{1}^2}{1+x_{1}x_{2}}+\frac{1+x_{2}^2}{1+x_{2}x_{3}}+\cdots+\frac{1+x_{n}^2}{1+x_{n}x_{1}}\geq n.\] simple use Of Titu lemma

we should prove that; \[\sum\frac{1}{1+x_ix_{i+1}}+\sum\frac{x_i^2}{1+x_ix_{i+1}}\ge n\]now apply titu we get: \[LH \ge \frac{n^2}{n+\sum x_ix_{i+1}}+\frac{\sum x_i^2}{n+\sum x_ix_{i+1}}\ge^? n\]which is obviously true because: \[\sum \frac{x_i}{2}+\frac{x_{i+1}}{2} \ge \sum x_ix_{i+1}\]

Woah this is pretty cool. By Cauchy $$(1+a^2)(1+b^2)\ge (1+ab)^2 \implies \dfrac{1+a^2}{1+ab} \ge \dfrac{1+ab}{1+b^2}$$So the equation turns into \begin{align*} 2 \left(\frac{1+x_{1}^2}{1+x_{1}x_{2}}+\frac{1+x_{2}^2}{1+x_{2}x_{3}}+\cdots+\frac{1+x_{n}^2}{1+x_{n}x_{1}} \right) &\ge \frac{1+x_{1}^2}{1+x_{1}x_{2}}+\frac{1+x_{2}^2}{1+x_{2}x_{3}}+\cdots+\frac{1+x_{n}^2}{1+x_{n}x_{1}} + \dfrac{1+x_1x_2}{1+x_2^2} + \dfrac{1+x_2x_3}{1+x_3^2} + \cdots + \dfrac{1+x_nx_1}{1+x_1^2} \\ &\ge 2n \end{align*}where the last inequality follows from AM-GM and we are done.

ARMO 2018 G11P2. For $n\in\mathbb Z_{\ge 2}$ and $x_1,\dots,x_n\in\mathbb R_+$ show that \[\sum_{k=1}^n\frac{1+x_k^2}{1+x_kx_{k+1}}\ge n.\] Solution. We use AM-GM to kill by \[\sum_{k=1}^n\frac{1+x_k^2}{1+x_kx_{k+1}}\ge\frac12\left(\sum_{k=1}^n\frac{1+x_k^2}{1+x_kx_{k+1}}+\sum_{k=1}^n\frac{1+x_kx_{k+1}}{1+x_{k+1}^2}\right)\ge n.\]

Solved with a hint (I can't do $n$-variable inequalities for some reason) It suffices to show that the product of the fractions is at least $1$, i.e. $$\prod_{i=1}^n (1+x_i^2) \geq \prod_{i=1}^n (1+x_ix_{i+1}),$$where $x_{n+1}=x_1$. Because we forget Cauchy-Schwarz exists we essentially use it without using it. For any distinct indices $i_1,\ldots,i_k$ such that $1 \leq i_j \leq n$ for all $1 \leq j \leq k$, we have the inequality $$x_{i_1}^2\ldots x_{i_k}^2+x_{i_1+1}^2\ldots x_{i_k+1}^2 \geq \frac{1}{2} x_{i_1}x_{i_1+1}\ldots x_{i_k}x_{i_k+1}.$$Summing this over all choices of $i_1,\ldots,i_k$ over all choices of $0 \leq k \leq n$ yields the expansion of the desired inequality, so we are done. $\blacksquare$

Straight AM-GM works! Notice that $$\sum_{i=1}^n \frac{1+x_i^2}{1+x_ix_{i+1}} \geq n \sqrt[n]{\frac{\prod_{i=1}^n (1+x_i)^2}{\prod_{i=1}^n (1+x_ix_{i+1})}}.$$But just multiply $(1+x_i^2)(1+x_{i+1}^2) \geq (1+x_ix_{i+1})^2$ cyclically to get the result.

calculus but in a weird way

Taking AM-GM on the whole expression gives \[\frac{1+x_{1}^2}{1+x_{1}x_{2}}+\frac{1+x_{2}^2}{1+x_{2}x_{3}}+\cdots+\frac{1+x_{n}^2}{1+x_{n}x_{1}} \ge n \sqrt[n]{\prod_{\text{cyc}} \frac{1+x_{1}^2}{1+x_{1}x_{2}}}\] so it suffices to prove \[\prod_{\text{cyc}} \frac{1+x_{1}^2}{1+x_{1}x_{2}} \ge 1\] which is easily shown as \[(1+x_i^2)(1+x_{i+1}^2) \ge (1+x_ix_{i+1})^2\] by Cauchy. Simply multiplying cyclically gives the desired result. $\square$

By AM-GM it is sufficient to prove $\prod_{\text{cyc}} (1 + x_1^2) \ge \prod_{\text{cyc}} (1 + x_1 x_2)$ which follows from multiplying the Cauchy Schwarz $\sqrt{(1 + x_1^2)(1 + x_2^2)} \ge 1 + x_1x_2$ cyclically.

By AM-GM, note that \begin{align*} \frac{1+x_{1}^2}{1+x_{1}x_{2}}+\frac{1+x_{2}^2}{1+x_{2}x_{3}}+\cdots+\frac{1+x_{n}^2}{1+x_{n}x_{1}} \geq n\sqrt[n]{\dfrac{\left(1+x_1^2\right)\left(1+x_2^2\right)\cdots \left(1+x_n^2\right)}{\left(1+x_1x_2\right)\left(1+x_2x_3\right)\cdots\left(1+x_nx_1\right)}} \end{align*}So, we want to show \begin{align*} \left(1+x_1^2\right)\left(1+x_2^2\right)\cdots \left(1+x_n^2\right) \geq \left(1+x_1x_2\right)\left(1+x_2x_3\right)\cdots\left(1+x_nx_1\right) \end{align*}Notice, \begin{align*} \left(1+x_1^2\right)\left(1+x_2^2\right) \geq \left(1+x_1x_2\right) \quad \iff \quad x_1^2 + x_2^2 \geq 2x_1x_2, \end{align*}which is clearly true by AM-GM. Multiplying cyclically yields the desired result.

Let the indices be under $\pmod n.$ By AM-GM, we have that $$\text{LHS} \geq n \sqrt[n]{\prod_{i=1}^{n} \frac{1+x_1^2}{1+x_1x_2}},$$so it suffices to show that $$\prod_{i=1}^{n} \frac{1+x_1^2}{1+x_1x_2} \geq 1.$$However, note that by Cauchy-Schwartz $$(x_i^2+1)(x_{i+1}^2+1) \geq (x_ix_{i+1}+1)^2,$$so multiplying all of these inequalities for $i=1, 2, \cdots, n$ and then taking the square root yields the desired result. QED