http://artofproblemsolving.com/community/c6h391671p2179241 Last post. For completeness' sake I'll post what I did in the paper (not verbatim). $$\frac{x}{y}+\frac{y}{z}+\frac{z}{x}-3 = \frac{(x-y)^2}{xy}+\frac{(z-x)(z-y)}{xz}$$Thus it suffices to prove that: $$\frac{(a-b)^2}{(a-1)(b-1)} \geq \frac{(a-b)^2}{(a+1)(b+1)}$$and $$\frac{(c-a)(c-b)}{(c-1)(a-1)} \geq \frac{(c-a)(c-b)}{(c+1)(a+1)}$$The first one is trivially true for $a, b, c > 1$. For the second one we can assume $\text{WLOG } c = min\{a,b,c\}$(since the inequality is cyclic) and then the inequality trivially follows. This method allows to generalize: $$\left(\frac{a}{b}\right)^n+\left(\frac{b}{c}\right)^n+\left(\frac{c}{a}\right)^n \ge \left(\frac{a+\lambda}{b+\lambda}\right)^n +\left(\frac{b+\lambda}{c+\lambda}\right)^n +\left(\frac{c+\lambda}{a+\lambda}\right)^n \text{ where }a,b,c,n,\lambda > 0$$

thewitness wrote: Let \(x,y,z\) be real numbers, each greater than \(1\). Prove that \(\dfrac{x+1}{y+1}+\dfrac{y+1}{z+1}+\dfrac{z+1}{x+1} \leq \dfrac{x-1}{y-1}+\dfrac{y-1}{z-1}+\dfrac{z-1}{x-1}\). $\frac{x-1}{y-1} - \frac{x+1}{y+1} = \frac{2(x-y)}{y^2-1}$, so it suffices to show that $$ \frac{x}{x^2-1} + \frac{y}{y^2-1} + \frac{z}{z^2-1} \leq \frac{x}{y^2-1} + \frac{y}{z^2-1} + \frac{z}{x^2-1}$$which is just rearrangement, as $x,y,z$ and $\frac{1}{x^2-1} , \frac{1}{y^2-1} , \frac{1}{z^2-1}$ are oppositely sorted.

My solution- It can be written as- $\frac{x-y}{y^2-1}+\frac{y-z}{z^2-1}+\frac{z-x}{x^2-1} \ge 0$. Assume WLOG $x \ge y \ge z$ Put $\frac{1}{z^2-1}=a$,$\frac{1}{y^2-1}=b$,$\frac{1}{x^2-1}=c$ Now $a \ge b \ge c$ We need to show $\frac{x-y}{y^2-1}+\frac{y-z}{z^2-1} \ge \frac{x-z}{x^2-1}$ or equivalently, $b(x-y) +a(y-z) \ge c(x-z)$ which is true by adding the following- $b(x-y) \ge c(x-y)$ and $a(y-z) \ge c(y-z)$

biomathematics wrote: thewitness wrote: Let \(x,y,z\) be real numbers, each greater than \(1\). Prove that \(\dfrac{x+1}{y+1}+\dfrac{y+1}{z+1}+\dfrac{z+1}{x+1} \leq \dfrac{x-1}{y-1}+\dfrac{y-1}{z-1}+\dfrac{z-1}{x-1}\). $\frac{x-1}{y-1} - \frac{x+1}{y+1} = \frac{2(x-y)}{y^2-1}$, so it suffices to show that $$ \frac{x}{x^2-1} + \frac{y}{y^2-1} + \frac{z}{z^2-1} \leq \frac{x}{y^2-1} + \frac{y}{z^2-1} + \frac{z}{x^2-1}$$which is just rearrangement as $x,y,z$ and $\frac{1}{x^2-1} , \frac{1}{y^2-1} , \frac{1}{z^2-1}$ are oppositely sorted. I did the same thing. Should you assume \(x \geq y \geq z\) or \(x \geq z \geq y\) and consider both cases to get that conclusion?

thewitness wrote: biomathematics wrote: thewitness wrote: Let \(x,y,z\) be real numbers, each greater than \(1\). Prove that \(\dfrac{x+1}{y+1}+\dfrac{y+1}{z+1}+\dfrac{z+1}{x+1} \leq \dfrac{x-1}{y-1}+\dfrac{y-1}{z-1}+\dfrac{z-1}{x-1}\). $\frac{x-1}{y-1} - \frac{x+1}{y+1} = \frac{2(x-y)}{y^2-1}$, so it suffices to show that $$ \frac{x}{x^2-1} + \frac{y}{y^2-1} + \frac{z}{z^2-1} \leq \frac{x}{y^2-1} + \frac{y}{z^2-1} + \frac{z}{x^2-1}$$which is just rearrangement as $x,y,z$ and $\frac{1}{x^2-1} , \frac{1}{y^2-1} , \frac{1}{z^2-1}$ are oppositely sorted. I did the same thing. Should you assume \(x \geq y \geq z\) or \(x \geq z \geq y\) and consider both cases to get that conclusion? No why? How does it matter?

div5252 wrote: My solution- It can be written as- $\frac{x-y}{y^2-1}+\frac{y-z}{z^2-1}+\frac{z-x}{x^2-1} \ge 0$. Assume WLOG $x \ge y \ge z$ Put $\frac{1}{z^2-1}=a$,$\frac{1}{y^2-1}=b$,$\frac{1}{x^2-1}=c$ Now $a \ge b \ge c$ We need to show $\frac{x-y}{y^2-1}+\frac{y-z}{z^2-1} \ge \frac{x-z}{x^2-1}$ or equivalently, $b(x-y) +a(y-z) \ge c(x-z)$ which is true by adding the following- $b(x-y) \ge c(x-y)$ and $a(y-z) \ge c(y-z)$ I just did the exact same thing

biomathematics wrote: thewitness wrote: Let \(x,y,z\) be real numbers, each greater than \(1\). Prove that \(\dfrac{x+1}{y+1}+\dfrac{y+1}{z+1}+\dfrac{z+1}{x+1} \leq \dfrac{x-1}{y-1}+\dfrac{y-1}{z-1}+\dfrac{z-1}{x-1}\). $\frac{x-1}{y-1} - \frac{x+1}{y+1} = \frac{2(x-y)}{y^2-1}$, so it suffices to show that $$ \frac{x}{x^2-1} + \frac{y}{y^2-1} + \frac{z}{z^2-1} \leq \frac{x}{y^2-1} + \frac{y}{z^2-1} + \frac{z}{x^2-1}$$which is just rearrangement as $x,y,z$ and $\frac{1}{x^2-1} , \frac{1}{y^2-1} , \frac{1}{z^2-1}$ are oppositely sorted. Cool!

biomathematics wrote: thewitness wrote: Let \(x,y,z\) be real numbers, each greater than \(1\). Prove that \(\dfrac{x+1}{y+1}+\dfrac{y+1}{z+1}+\dfrac{z+1}{x+1} \leq \dfrac{x-1}{y-1}+\dfrac{y-1}{z-1}+\dfrac{z-1}{x-1}\). $\frac{x-1}{y-1} - \frac{x+1}{y+1} = \frac{2(x-y)}{y^2-1}$, so it suffices to show that $$ \frac{x}{x^2-1} + \frac{y}{y^2-1} + \frac{z}{z^2-1} \leq \frac{x}{y^2-1} + \frac{y}{z^2-1} + \frac{z}{x^2-1}$$which is just rearrangement as $x,y,z$ and $\frac{1}{x^2-1} , \frac{1}{y^2-1} , \frac{1}{z^2-1}$ are oppositely sorted. I did EXACTLY the same thing!!!!!!!!!!!!!!

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thewitness wrote: Let \(x,y,z\) be real numbers, each greater than \(1\). Prove that \(\dfrac{x+1}{y+1}+\dfrac{y+1}{z+1}+\dfrac{z+1}{x+1} \leq \dfrac{x-1}{y-1}+\dfrac{y-1}{z-1}+\dfrac{z-1}{x-1}\). The inequality is equivalent to Let $x,y,z > 0$ . Prove that $$\dfrac{x+2}{y+2}+\dfrac{y+2}{z+2}+\dfrac{z+2}{x+2} \leq \dfrac{x}{y}+\dfrac{y}{z}+\dfrac{z}{x}$$See also here https://artofproblemsolving.com/community/c6h427801p5449971

this is where geniuses hang out, huh?

i have solved it using rearrangement in the paper. but can i use rearrangement without giving its proof?how much marks can i expect for a solution without a proof for rearrangement?

Zeref123 wrote: i have solved it using rearrangement in the paper. but can i use rearrangement without giving its proof?how much marks can i expect for a solution without a proof for rearrangement? I don't really expect any marks to be deducted for that honestly.

I also did it using rearrangement

With the substitutions $x=1+a, y=1+b, z=1+c$, the given inequality is equivalent to $$\sum_{cyc} \frac{a}{b} \geq \sum_{cyc} \frac{a+2}{b+2} \iff \sum_{cyc} \frac{a-b}{b(b+2)} \geq 0 ...(1)$$for $a, b, c \in \mathbb{R}_0$. Now, if $a \geq b \geq c$, $(1)$ is equivalent to $$(a-b)(\frac{1}{b(b+2)} - \frac{1}{a(a+2)}) + (b-c)(\frac{1}{c(c+2)} - \frac{1}{a(a+2)}) \geq 0,$$whereas if $a \geq c \geq b$, for which $(1)$ is equivalent to $$(c-b)(\frac{1}{b(b+2)} - \frac{1}{c(c+2)}) + (a-c)(\frac{1}{b(b+2)} - \frac{1}{a(a+2)}) \geq 0$$both of which are obvious. Alternatively: We just assume wlog that $c=min(a,b,c)$ and observe that $(1)$ is equivalent to $$\frac{(a-b)^2(a+b+2)}{ab(a+2)(b+2)} + \frac{(a-c)(b-c)(a+c+2)}{ac(a+2)(c+2)} \geq 0$$which is again obvious.

Using the rearrangement inequality, can one conclude this? For $a,b,c \in R^+$ $$\sum_{cyc}a^{3}c^{2} \geq \sum_{cyc}a^{2}bc^{2}$$

Someone pls reply ASAP

What do you guys think should be the cutoff for this year....I solved three questions perfectly

It seems that all regions had the same paper

https://artofproblemsolving.com/community/c6t321f6h1349732_inequalities_with_degree_gt_2 Already posted much earlier.

Ignore....

@above, I don't think you can multiply the numerator and denominator by (x-y),(y-z),etc. as x,y,z are NOT guaranteed to be distinct.

The proof is evidently wrong.

MathematicalPhysicist wrote: i did it by cauchy schwartz. Can you please tell me how did you use cs here

Assume that $z=\min\lbrace x,y,z\rbrace$. We can do it since inequality is cyclic. Then $(y-z)(x-z)\ge 0$. Hence $$\frac{\text{LHS}-\text{RHS}}{2}=\sum_{cyc}\frac{x-y}{y^2-1}=\frac{x-y}{y^2-1}-\frac{x-y}{x^2-1}+\frac{y-z}{z^2-1}-\frac{y-z}{x^2-1}=$$$$=\frac{(x-y)^2(x+y)}{(x^2-1)(y^2-1)}+\frac{(y-z)(x-z)(x+z)}{(x^2-1)(z^2-1)}\ge 0$$Remark You could also consider $z$ as the greatest of the three numbers and above solution still works.

thewitness wrote: Let \(x,y,z\) be real numbers, each greater than \(1\). Prove that \(\dfrac{x+1}{y+1}+\dfrac{y+1}{z+1}+\dfrac{z+1}{x+1} \leq \dfrac{x-1}{y-1}+\dfrac{y-1}{z-1}+\dfrac{z-1}{x-1}\). Subtract LHS from RHS. We are left to prove, after rearranging some terms, that $$ \frac{x}{x^2-1} + \frac{y}{y^2-1} + \frac{z}{z^2-1} \leq \frac{x}{y^2-1} + \frac{y}{z^2-1} + \frac{z}{x^2-1}$$Now observe that this is trivial by rearrangement, as $x,y,z$ are sorted opposite to $\frac{1}{x^2-1},\frac{1}{y^2-1},\frac{1}{z^2-1}$ I wish i had known Rearrangement that day and I would have nailed it

(x+1)/(y+1) = 1 + {(x-y)/(y+1)} and also (x-1)/(y-1) = 1 + {(x-y)/(y-1)} hence it's sufficient to prove that (x-y)/(y+1) + (y-z)/(z+1) + (z-x)/(x+1) <= (x-y)/(y-1) + (y-z)/(z-1) + (z-x)/(x-1) but from rearrangement inequality , we get y/(y-1) + z/(z-1) + x/(x-1) <= x/(y-1) + y/(z-1) + z/(x-1) y/(y+1) + z/(z+1) + x/(x+1) <= x/(y+1) + y/(z+1) + z/(x+1) some simplification gives the desired result

WLOG assume $x-1=a,y-1=b,z-1=c$ clearly $a,b,c$ are positive reals. For contrary assume $\sum \frac{a+2}{b+2} >\sum \frac{a}{b}$. $or,\sum \frac{a+2}{b+2}-\frac{a}{b}>0$. $or,\sum \frac{2(b-a)}{b(b+2)} >0$. WLOG assume $a>b >c$. $or,0<\sum \frac{2(b-a)}{b(b+2)} <\frac{b-a+a-c+c-b}{c(c+2)}=0$. Contradiction . $or,\sum{a+2}{b+2} \le \sum{a}{b}$. Which follows conclusion !!

Another solution using rearrangement inequality. as we have assume before $or,\sum \frac{2(b-a)}{b(b+2)} >0$. $\sum \frac{b}{b(b+2)} > \sum\frac{a}{b(b+2)}$. But it is contradiction since $\sum{b}{b(b+2)}\le \sum{a}{b(b+2)}$.

Here's a solution, Let $x \ge y \ge z$ We need to prove that, $\frac{x-y}{y^2-1} + \frac{y-z}{z^2-1}+\frac{z-x}{x^2-1} \ge 0$ or, $\frac{x-y}{y^2-1} + \frac{y-z}{z^2-1} \ge \frac{x-z}{x^2-1}$ or, $\frac{x-y}{y^2-1} + \frac{y-z}{z^2-1} \ge \frac{x-y}{x^2-1} + \frac{y-z}{x^2-1}$ Which is obvious, as $\frac{1}{x^2-1} \le \frac{1}{y^2-1} \le \frac{1}{z^2-1}$

arulxz wrote: Let $x-1=a,y-1=b,z-1=c$. By titu, $$\sum_{cyc} \frac{(a-b)^2}{b(b+2)(a-b)} \geq \frac{(a-b+b-c+c-a)^2}{\sum_{cyc} b(b+2)(a-b)} = 0$$ Since $$2\sum_{cyc} \frac{a-b}{b(b+2)} =\sum_{cyc} \frac{x-1}{y-1} - \frac{x+1}{y+1}$$we are done. This solution is wrong as you can apply titu's lemma only when the denominator is also positive

Claim : If $x\ge y$ and $p\ge 0$ Then $\frac{x-p}{y-p}\ge\frac{x}{y}\ge\frac{x+p}{y+p}$ Proof: we don't know if $\frac{x}{y}\ge\frac{x+p}{y+p}$. So let $\frac{x}{y}\lesseqgtr\frac{x+p}{y+p}$ $\implies xy+xp\lesseqgtr xy+py\implies x\lesseqgtr y\implies x>y$ so we get $\frac{x}{y}\ge\frac{x+p}{y+p}$ from $\frac{x}{y}\lesseqgtr\frac{x+p}{y+p}$ Solution: $\sum\frac{x-p}{y-p}\ge\sum\frac{x}{y}\ge\sum\frac{x+p}{y+p}$ set $p=1$ we get the the result

amitwa.exe wrote: Claim : If $x\ge y$ and $p\ge 0$ Then $\frac{x-p}{y-p}\ge\frac{x}{y}\ge\frac{x+p}{y+p}$ Proof: we don't know if $\frac{x}{y}\ge\frac{x+p}{y+p}$. So let $\frac{x}{y}\lesseqgtr\frac{x+p}{y+p}$ $\implies xy+xp\lesseqgtr xy+py\implies x\lesseqgtr y\implies x>y$ so we get $\frac{x}{y}\ge\frac{x+p}{y+p}$ from $\frac{x}{y}\lesseqgtr\frac{x+p}{y+p}$ Solution: $\sum\frac{x-p}{y-p}\ge\sum\frac{x}{y}\ge\sum\frac{x+p}{y+p}$ set $p=1$ we get the the result I think that this solution is wrong. Why? Because we assume $x \geq y$ for the following inequality, but what about $\frac{z}{x}$. This does not satisfy this.

substitute $x=1+a$, $y=1+b$, and $z=1+c$. The given inequality reduces to: $\frac{a}{b(b+2)}+\frac{b}{c(c+2)}+\frac{c}{a(a+2)}\geq \frac{b}{b(b+2)}+\frac{c}{c(c+2)}+\frac{a}{a(a+2)}$, which is trivial by the rearrangement inequality