wavelet3000 wrote: $x$, $y$ and $z$ are positive real numbers. Prove the inequality \[\frac{x+1}{y+1}+\frac{y+1}{z+1}+\frac{z+1}{x+1}\leq\frac{x}{y}+\frac{y}{z}+\frac{z}{x}.\] (Authors: A. Khrabrov, B. Trushin) Since $\frac{x}{y}-\frac{x+1}{y+1}=\frac{x-y}{y(y+1)},$ the inequality can be written as $\frac{x-y}{y(y+1)}+\frac{y-z}{z(z+1)}+\frac{z-x}{x(x+1)} \ge 0,$ or $\left[ \frac{x-y}{y(y+1)}+1\right]+\left[\frac{y-z}{z(z+1)}+1\right]+\left[\frac{z-x}{x(x+1)}+1\right] \ge 3.$ This rewrites as $\frac{x+y^2}{y(y+1)}+\frac{y+z^2}{z(z+1)}+\frac{z+x^2}{x(x+1)} \ge 3.$ Now, using the Cauchy-Schwarz inequality, we get $x+y^2 =\frac{(x+y^2)(1+x)}{1+x} \ge \frac{x(1+y)^2}{1+x}.$ Therefore, $\frac{x+y^2}{y(y+1)} \ge \frac{x(1+y)}{y(1+x)}.$ Finally, we come up with the inequality $\frac{x(1+y)}{y(1+x)}+\frac{y(1+z)}{z(1+y)}+\frac{z(1+x)}{x(1+z)} \ge 3,$ which is true according to the AM-GM inequality.

can_hang2007 wrote: Since $\frac{x}{y}-\frac{x+1}{y+1}=\frac{x-y}{y(y+1)},$ the inequality can be written as $\frac{x-y}{y(y+1)}+\frac{y-z}{z(z+1)}+\frac{z-x}{x(x+1)} \ge 0,$ we can also use rearrangement here, no? note that $(x,y,z)$ and $\left(\frac1{x(x+1)},\frac1{y(y+1)},\frac1{z(z+1)}\right)$ are oppositely sorted.

Johan Gunardi wrote: we can also use rearrangement here, no? note that $(x,y,z)$ and $\left(\frac1{x(x+1)},\frac1{y(y+1)},\frac1{z(z+1)}\right)$ are oppositely sorted. I know that solution, but I think it is over the level of grade 9.

your way to apply cauchy-schwarz is way beyond 9 grade level. another solution is using the identity $\frac{a}b+\frac{b}c+\frac{c}a=3+\frac{(a-b)^2}{ab}+\frac{(a-c)(b-c)}{ac}$.

Johan Gunardi wrote: your way to apply cauchy-schwarz is way beyond 9 grade level. Sorry, but for Olympiad at secondary school, I dont think it is over grade 9 (at least in Vietnam, it is).

This following one is also true. \[a,b,c,d > 0\] \[\sum\limits_{cyc} {\sqrt {\frac{{{a^2} + 2011}}{{{b^2} + 2011}}} } \le \sum\limits_{cyc} {\frac{a}{b}} \]

You can see here generalization and my proof.

MathUniverse wrote: You can see here generalization and my proof. I have an interesting proof,with Cauchy-Schwarz and

I think that is over 9th grade. To finish my proof for generalization, we just need to note

Also, I've found a proof for your inequality only using cauchy-schwarz.

Really nice soluyion,Mathuniverse! But I think you don't need Cauchy-Schwarz in your proof... Note that:\[x + \frac{1}{x} \ge y + \frac{1}{y} \Leftrightarrow (x - y)(xy - 1) \ge 0\] Let \[x = \frac{a}{b},y = \sqrt {\frac{{{a^2} + 2011}}{{{b^2} + 2011}}} \] \[a > b \Rightarrow x > y > 1;a \le b \Rightarrow x \le y \le 1\] So,\[\frac{a}{b} + \frac{b}{a} \ge \sqrt {\frac{{{a^2} + 2011}}{{{b^2} + 2011}}} + \sqrt {\frac{{{b^2} + 2011}}{{{a^2} + 2011}}} \] P.S:this method also works well for the original inequality I'll post my proof later

$x$, $y$ and $z$ are positive real numbers. Prove the inequality \[\sqrt{\frac{x+1}{y+1}}+\sqrt{\frac{y+1}{z+1}}+\sqrt{\frac{z+1}{x+1}}\leq\sqrt{\frac{x}{y}}+\sqrt{\frac{y}{z}}+\sqrt{\frac{z}{x}}.\]

Let $a,b,c$ be positive real numbers. For $0\leq\alpha< \beta$ , prove that \[\frac{a+\alpha}{b+\alpha}+\frac{b+\alpha}{c+\alpha}+\frac{c+\alpha}{a+\alpha} \geq \frac{a+ \beta}{b+ \beta}+\frac{b+ \beta}{c+ \beta}+\frac{c+ \beta}{a+ \beta}.\]

wavelet3000 wrote: $x$, $y$ and $z$ are positive real numbers. Prove the inequality \[\frac{x+1}{y+1}+\frac{y+1}{z+1}+\frac{z+1}{x+1}\leq\frac{x}{y}+\frac{y}{z}+\frac{z}{x}.\](Authors: A. Khrabrov, B. Trushin) Obvious by full expanding and Muirhead

sqing wrote: Let $a,b,c$ be positive real numbers. For $0\leq\alpha< \beta$ , prove that \[\frac{a+\alpha}{b+\alpha}+\frac{b+\alpha}{c+\alpha}+\frac{c+\alpha}{a+\alpha} \geq \frac{a+ \beta}{b+ \beta}+\frac{b+ \beta}{c+ \beta}+\frac{c+ \beta}{a+ \beta}.\] Just apply the original inequality on the positive numbers $x=\frac{a+\alpha}{\beta-\alpha}, y=\frac{b+\alpha}{\beta-\alpha}$ and $z=\frac{c+\alpha}{\beta-\alpha}$.

wavelet3000 wrote: $x$, $y$ and $z$ are positive real numbers. Prove the inequality \[\frac{x+1}{y+1}+\frac{y+1}{z+1}+\frac{z+1}{x+1}\leq\frac{x}{y}+\frac{y}{z}+\frac{z}{x}.\](Authors: A. Khrabrov, B. Trushin) http://www.artofproblemsolving.com/community/c6h604980p5626079: Prove that if $a,b,c$ are positive numbers with $abc=1$, then $$\frac{a+1}{b+1}+\frac{b+1}{c+1}+\frac{c+1}{a+1}\leq a+b+c \leq\frac{a}{b} +\frac{b}{c} + \frac{c}{a} .$$

DC93 wrote: This following one is also true. \[a,b,c,d > 0\]\[\sum\limits_{cyc} {\sqrt {\frac{{{a^2} + 2011}}{{{b^2} + 2011}}} } \le \sum\limits_{cyc} {\frac{a}{b}} \] For positives $a$, $b$ and $c$ prove that: $$\frac ab+\frac bc+\frac ca \ge \sqrt{\frac{a^2+2011}{b^2+2011}}+\sqrt{\frac{b^2+2011}{c^2+2011}}+\sqrt{\frac{c^2+2011}{a^2+2011}}$$$$\iff$$$$\frac ab+\frac bc+\frac ca \ge \sqrt{\frac{a^2+1}{b^2+1}}+\sqrt{\frac{b^2+1}{c^2+1}}+\sqrt{\frac{c^2+1}{a^2+1}}$$

wavelet3000 wrote: $x$, $y$ and $z$ are positive real numbers. Prove the inequality \[\frac{x+1}{y+1}+\frac{y+1}{z+1}+\frac{z+1}{x+1}\leq\frac{x}{y}+\frac{y}{z}+\frac{z}{x}.\](Authors: A. Khrabrov, B. Trushin) https://artofproblemsolving.com/community/c6h506574p8907644: $$\left(\frac{a}{b}\right)^n+\left(\frac{b}{c}\right)^n+\left(\frac{c}{a}\right)^n \ge \left(\frac{a+1}{b+1}\right)^n +\left(\frac{b+1}{c+1}\right)^n +\left(\frac{c+1}{a+1}\right)^n . $$

sqing wrote: Prove that if $a,b,c$ are positive numbers with $abc=1$, then $$\frac{a+1}{b+1}+\frac{b+1}{c+1}+\frac{c+1}{a+1}\leq a+b+c \leq\frac{a}{b} +\frac{b}{c} + \frac{c}{a} .$$ Prove that if $a,b,c$ are positive numbers with $abc=1$, then $$\frac{a}{b}+\frac{a}{c}+\frac{b+c}{a}\geq a+b+c+1.$$

wavelet3000 wrote: $x$, $y$ and $z$ are positive real numbers. Prove the inequality \[\frac{x+1}{y+1}+\frac{y+1}{z+1}+\frac{z+1}{x+1}\leq\frac{x}{y}+\frac{y}{z}+\frac{z}{x}.\](Authors: A. Khrabrov, B. Trushin) See here for more : https://artofproblemsolving.com/community/c6t321f6h1349732_inequalities_with_degree_gt_2

Easy problem

$ \sum_{cyc} \frac{x}{y} \geq \sum_{cyc} \frac{x+1}{y+1} $ $ \sum_{cyc} \left ( \frac{x}{y} - \frac{x+1}{y+1} \right ) = \sum_{cyc} \frac{x-y}{y(y+1)} \geq 0 $ Dividing by y, $ (*!) $ $ \sum_{cyc} \frac {x-y}{y+1} \geq 0 $ Substitute $ \begin{cases} a = x+1 \\ b=y+1 \\ c=z+1 \end{cases} $ $ [ \frac {c-a}{a} +`1 ] + [ \frac {a-b}{b} + 1 ] + [ \frac {b-c}{c} + 1 ] \geq 3 $ $ \sum_{cyc} \frac {a}{b} \geq 3 $ Which is very easy by AM-GM: $ \frac { \frac {a}{b} + \frac {b}{c} + \frac {c}{a} } {3} \geq \sqrt[3]{ \frac {abc}{abc} } $ $ \Rightarrow $ $ \sum_{cyc} \frac {a}{b} \geq 3 \sqrt[3]{1} = 3 $ EDIT: Not too sure if I can just take out a y-term in the cyclic sum at $ (*!) $

sqing wrote: $x$, $y$ and $z$ are positive real numbers. Prove the inequality \[\sqrt{\frac{x+1}{y+1}}+\sqrt{\frac{y+1}{z+1}}+\sqrt{\frac{z+1}{x+1}}\leq\sqrt{\frac{x}{y}}+\sqrt{\frac{y}{z}}+\sqrt{\frac{z}{x}}.\] $$\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\ge \sqrt{\frac{1+a^2}{1+b^2}}+\sqrt{\frac{1+b^2}{1+c^2}}+\sqrt{\frac{1+c^2}{1+a^2}}$$North Korea Team Selection Test 2016： Let $a_1,a_2,\cdots ,a_n>0$.Prove that$$\sum_{k=1}^{n}\frac{a_{k+1}}{a_k}\geq \sum_{k=1}^{n}\sqrt{\frac{a^2_{k+1}+1}{a^2_k+1}}$$Where $a_{n+1}=a_1$ Let $a,b,c$ be positive real numbers. Prove that $$\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\geq \sqrt{\frac{a^{2}+c^{2}}{b^{2}+c^{2}}}+\sqrt{\frac{b^{2}+a^{2}}{c^{2}+a^{2}}}+\sqrt{\frac{c^{2}+b^{2}}{a^{2}+b^{2}}}$$

Let $a,b,c$ be positive real numbers. Prove that $$\sqrt[3]{\frac {a}{b}}+\sqrt[3]{\frac {b}{c}}+\sqrt[3]{\frac {c}{a}}\geq \sqrt[3]{\frac {a + b}{a + c}}+\sqrt[3]{\frac {b + c}{b + a}}+\sqrt[3]{\frac {c + a}{c + b}}$$Let $a,b,c$ be positive real numbers. Prove that \[\sqrt{\frac{a}b}+\sqrt{\frac{b}c}+\sqrt{\frac{c}a} \geq\sqrt{\frac{a+b}{a+c}}+\sqrt{\frac{b+c}{b+a}}+ \sqrt{\frac{c+a}{c+b}}.\]here Let $a,b,c$ be positive real numbers such that $a+b+c\leq 1. $ Prove that$$\frac{a}{b^2}+\frac{b}{c^2}+\frac{c}{a^2}+1-a-b-c\geq (\frac{a}{b}+\frac{b}{c}+\frac{c}{a}+1-a-b-c)^2.$$

AmSm_9 wrote: EDIT: Not too sure if I can just take out a y-term in the cyclic sum at $ (*!) $ No, you can't.

Oops I messed up

sqing wrote: North Korea Team Selection Test 2016： Let $a_1,a_2,\cdots ,a_n>0$.Prove that$$\sum_{k=1}^{n}\frac{a_{k+1}}{a_k}\geq \sum_{k=1}^{n}\sqrt{\frac{a^2_{k+1}+1}{a^2_k+1}}$$Where $a_{n+1}=a_1$ MariusStanean wrote: My solution: let $a_i=\tan\alpha_i$ where $\alpha_i\in(0,\pi/2)$. Hence, we need to prove that $$\sum_{i=1}^n\frac{\sin\alpha_{i+1}}{\sin\alpha_i}\cdot\frac{\cos\alpha_i}{\cos\alpha_{i+1}}\ge \sum_{i=1}^n\frac{\cos\alpha_i}{\cos\alpha_{i+1}},$$or $$\sum_{i=1}^n\frac{\cos\alpha_i}{\cos\alpha_{i+1}}\cdot\frac{\sin\alpha_{i+1}-\sin\alpha_i}{\sin\alpha_i}\ge 0,$$or $$\sum_{i=1}^n\frac{\sin\alpha_{i+1}-\sin\alpha_i}{\sin\alpha_i}+\sum_{i=1}^n\frac{\cos\alpha_i-\cos\alpha_{i+1}}{\cos\alpha_{i+1}}\cdot\frac{\sin\alpha_{i+1}-\sin\alpha_i}{\sin\alpha_i}\ge 0$$which is obviously true, first sum from AM-GM Inequality and second sum from monotony. https://artofproblemsolving.com/community/c6h1566036p9594558

Let $a_1,a_2,\cdots ,a_n $ are positive real numbers such that $|a_i-a_j|\leq 1 $ $(1\leq i<j\leq n).$. Prove that$$\frac{a_1}{a_2}+\frac{a_2}{a_3}+\cdots+\frac{a_{n-1}}{a_n}+\frac{a_n}{a_1}\geq \frac{a_2+1}{a_1+1}+ \frac{a_3+1}{a_2+1}+\cdots +\frac{a_n+1}{a_{n-1}+1}+\frac{a_1+1}{a_n+1}.$$p/6232480389

Let $0<a\leq b\leq c $ . Prove that $$\frac{1}{2}\left(\frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}+3\right)\geq \frac{a}{b}+\frac{b}{c}+\frac{c}{a}\geq\frac{b}{a}+\frac{c}{b}+\frac{a}{c}\geq \frac{b+1}{a+1}+\frac{c+1}{b+1}+\frac{a+1}{c+1}$$