A less tricky way for less trickier people.

Note that by Cauchy-Schwarz inequaliy, we get $\frac{1}{x^2-1}+\frac{1}{x+1}+\frac{1}{x+1}\ge\frac{9}{(x+1)^2}$ And by $\frac{9}{(x+1)^2}\ge\frac{9}{x+1}-\frac{9}{4}$ Add them together we get, $\frac{1}{x^2-1}+\frac{9}{4}\ge\frac{7}{x+1}$ So, \begin{align*}\sum_{cyc} \frac{1}{x+1}&\le\frac{1}{7}(\sum_{cyc}(\frac{1}{x^2-1}+\frac{9}{4}))\\&=1\end{align*}

Note that by Cauchy-Schwarz inequality, we get $\frac{1}{x-1}+\frac{4}{2} \ge \frac{9}{x+1}\Rightarrow \boxed{\frac{1}{x-1}\ge\frac{9}{x+1}-2}$ By Chebychev inequality, we get $3\sum_{cyc}\frac{1}{x^2-1}\ge(\sum_{cyc}\frac{1}{x-1})(\sum_{cyc}\frac{1}{x+1})$ So, \begin{align*}3&\ge(\sum_{cyc}\frac{1}{x-1})(\sum_{cyc}\frac{1}{x+1})\\&\ge(\sum_{cyc}(\frac{9}{x+1}-2))(\sum\frac{1}{x+1})\end{align*}Let $S=\sum_{cyc}\frac{1}{x+1}$ from above hence, ${3\ge(9S-6)S}\Rightarrow \boxed{3(S-1)(3S+1)\le 0}$ It's trivial that $S\ge0$. So $S\le1\Rightarrow\boxed{\sum_{cyc}\frac{1}{x+1}\le1}$ as we want.

Assume that \(x,y,z > 1\), or else the condition doesn't make sense. The condition can be rewritten as \[\frac{x^2}{x^2-1}+\frac{y^2}{y^2-1}+\frac{z^2}{z^2-1}=4.\] The desired inequality can be rewritten as \begin{align*} \frac{x-1}{x^2-1}+\frac{y-1}{y^2-1}+\frac{z-1}{z^2-1}&\le \frac{1}{x^2-1} +\frac{1}{y^2-1}+\frac{1}{z^2-1}\\ \frac{x}{x^2-1}+\frac{y}{y^2-1}+\frac{z}{z^2-1}&\le \frac{2}{x^2-1} +\frac{2}{y^2-1}+\frac{2}{z^2-1} \end{align*}The right hand side is just \(2\). Note that for \(x,y,z > 1\) we should have \(x/(x^2-1)>0\), and the same for the other variables. Thus, by Cauchy, \[\frac{x}{x^2-1}+\frac{y}{y^2-1}+\frac{z}{z^2-1} \le \sqrt{\left(\frac{1}{x^2-1}+\frac{1}{y^2-1}+\frac{1}{z^2-1}\right)\left(\frac{x^2}{x^2-1}+\frac{y^2}{y^2-1}+\frac{z^2}{z^2-1} \right)}\] The right hand side is equal to \(2\), as desired.

Let $\frac{1}{x^2-1}=a$, with $a+b+c=1$ The inequality becomes $\sum_{ }^{ }\frac{\sqrt{a}}{\sqrt{a}+\sqrt{1+a}}\le1$ By the TLM, $\frac{\sqrt{x}}{\sqrt{1+x}+\sqrt{x}}\le\frac{1}{4}x+\frac{1}{4}$ with equality at $\frac{1}{3}$ and the result follows

No-Caushy approach: Let $a = \frac{1}{x+1}$ and similarly for $b$ and $c$, and let $s = a +b + c$ We now have $\frac{1}{x-1} = \frac{1}{\frac{1}{a}-2} = \frac{a}{1-2a}$ Next, let $f(x) = \frac{x^2}{1-2x}$, so that we have $f(a) + f(b)+f(c) = 1$. We get that $f$ i convex on $(0,1/2)$ and also from $x>1$ we have $a<1/2$, now we can use Jensen: $1 = f(a)+f(b)+f(c) \geq 3f(\frac{a+b+c}{3}) = \frac {s^2}{3-2s} \iff 3-2s \geq s^2 \iff s^2+2s-3 \leq 0 \iff (s-1)(s+3) \leq 0$ so we get $s \leq 1$

No cauchy or jensen, plain old AM GM. Let $a=\frac{1}{x^2-1}$ then we have $a+b+c=1$ and the inequality reduces to $\displaystyle\Sigma\frac{\sqrt a}{\sqrt{a+1}+\sqrt{a}}\leq 1$ Now rationalizing the denominator and simplifying we get $\iff \Sigma \sqrt{(a+1)a} \leq 2 \iff \Sigma \sqrt{\frac{(a+1)}{4}a} \leq 1$ But by AM-GM , $\sqrt{\frac{(a+1)}{4}a}\leq\frac{a+\frac{a+1}{4}}{2}$ Adding up we get $ \sqrt{\frac{(a+1)}{4}a} \leq 1$ Hence proved, Equality iff $a=b=c=\frac{1}{3}$.

The problem gives us \[ \sum \frac 1{x^2-1} - \frac 13 = \sum \frac{4-x^2}{3(x^2-1)} = 0 \iff \sum \frac{4-x^2}{x^2-1} = \sum \frac{2-x}{x+1} \cdot \frac{2+x}{x-1} = 0 \]and asks of us \[ \sum \frac 1{x+1} - \frac 13 =\sum \frac{2-x}{3(x+1)} \leq 0 \iff \sum \frac{2-x}{x+1}\leq 0. \]Without loss of generality impose the ordering $x\leq y \leq z$, such that $\left\{\frac{2-x}{x+1} = \frac{3}{x+1} - 1\right\}$ and $\left\{\frac{2+x}{x-1} =1+\frac{3}{x-1}\right\}$ are similarly ordered. Thus by Tschebyscheff , \[ 0 = \sum \frac{2-x}{x+1}\cdot \frac{2+x}{x-1} \geq \left(\sum \frac{2-x}{x+1}\right) \left(\sum \frac{2+x}{x-1}\right). \]The right factor is clearly positive, so the left factor is nonpositive. $\Box$

Letting $\frac{1}{x^2-1} = x'$ and similar for $y'$ and $z'$, we have $x'+y'+z'=1$ and letting $f(n)=\frac{1}{\sqrt{\frac{1+a}{a}}+1}$, we have that we need to prove $f(x')+f(y')+f(z') \le 1$ given that $x', y', z'$ are positive numbers. Taking its first derivative, we have that $$\frac{d}{dn}\left( \frac{1}{\sqrt{\frac{1+a}{a}}+1} \right) = \frac{\sqrt{\frac{n+1}{n}}}{2n(n+1)\left(\sqrt{\frac{n+1}{n}}+1\right)^2}$$which means that after taking the derivative of that, we see the second derivative is merely $$\frac{d}{dn}\left( \frac{\sqrt{\frac{n+1}{n}}}{2n(n+1)\left(\sqrt{\frac{n+1}{n}}+1\right)^2} \right) = \frac{\sqrt{\frac{n+1}{n}}}{4n(n+1)^2}$$which is always negative over its domain which includes all positive numbers, meaning that $f$ is concave down, so as a result, $$f(x')+f(y')+f(z') \le 3f\left(\frac13\right) = 3 \cdot \frac13 = 1$$which proves our desired inequality.

By Cauchy-Schwarz, we have \[\left(\sum \frac{1}{x^2-1}\right)\left(\sum \frac{x-1}{x+1}\right) \ge \left(\sum \frac{1}{x+1}\right)^2.\] Then if we let $A = \sum \frac{1}{x+1}$, we have \[3-2A \ge A^2\] which rearranges to \[(A-1)(A+3) \le 0.\] Since $A$ is clearly positive, we must have $A \le 1$, as desired. $\square$

From where do you get this? \[3-2A \ge A^2\]

\[\left(\sum \frac{1}{x^2-1}\right)\left(\sum \frac{x-1}{x+1}\right) = \sum \left( 1 - \frac{2}{x+1} \right) = 3 - 2A \]

Evan Chen made a typo. It is Po-Shen-Loh

Um, no, he didn't. Po-Ru Loh is a real person and I believe won 1999 Countdown round in MC Nats.

Notice that the given condition is equivalent to $$\sum_{cyc}\frac{1}{x-1}=2+\sum_{cyc}\frac{1}{x+1}$$By C-S $$\big(\sum_{cyc}\frac{1}{x^2-1}\big)\big(\sum_{cyc}\frac{x-1}{x+1}\big) \geq \big(\sum_{cyc}\frac{1}{x+1}\big)^2$$$$3-2\sum_{cyc}\frac{1}{x+1}\geq \big( \sum_{cyc} \frac{1}{x+1} \big)^2$$Let $S=\sum_{cyc}\frac{1}{x+1}$. Then $$0 \geq S^2+2S-3$$$$0 \geq (S+3)(S-1)$$Thus $1\geq S$, as desired. $\square$

By AM-GM, $\frac{1}{x^2-1}+\frac{x-1}{x+1} \ge \frac{2}{x+1}$. $\sum (\frac{1}{x^2-1}+(1- \frac{2}{x+1})) \ge \sum \frac{2}{x+1} \Longrightarrow 1 \ge \sum \frac{1}{x+1}$

Claim: $\frac{1}{x+1}\leq \frac{1}{4(x^2-1)}+\frac{1}{4}$ Proof: Expanding: \[x^2-1+1-4x+4\geq 0,\]\[(x-2)^2\geq 0,\]which is true by Trivial Inequality $\square$ Now, let $f(x)=\frac{1}{x+1}$. Therefore: \[f(x)+f(y)+f(z)\leq \sum_{\text{cyc}}\frac{1}{4(x^2-1)}+\frac{1}{4}=1,\]as desired $\blacksquare$

very cool problem By Cauchy, \begin{align*} \left(\frac1{x^2-1}+\frac1{y^2-1}+\frac1{z^2-1}\right)\left(\frac{x-1}{x+1}+\frac{y-1}{y+1}+\frac{z-1}{z+1}\right) &\ge \left(\frac1{x+1}+\frac1{y+1}+\frac1{z+1}\right)^2, \end{align*}so by manipulation and substituting known values, \begin{align*} 3-2\left(\frac1{x-1}+\frac1{y-1}+\frac1{z-1}\right) \ge \left(\frac1{x-1}+\frac1{y-1}+\frac1{z-1}\right)^2, \end{align*}and solving the quadratic gives the desired result.

seems to be equivalent to some of the above writeups (haven't looked throughly), but checking this for errors would be appreciated Notice that $\tfrac{1}{x^2-1}$ can be written as $\tfrac{1}{(x+1)^2}/\tfrac{x-1}{x+1}$. Use Titu’s Lemma: \[1=\frac{\frac{1}{(x+1)^2}}{\frac{x-1}{x+1}}+\frac{\frac{1}{(y+1)^2}}{\frac{y-1}{y+1}}+\frac{\frac{1}{(z+1)^2}}{\frac{z-1}{z+1}}\ge \frac{\left(\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}\right)^2}{\frac{x-1}{x+1}+\frac{y-1}{y+1}+\frac{z-1}{z+1}}=\frac{\left(\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}\right)^2}{3-2\left(\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}\right)}.\]Letting $s=\tfrac{1}{x+1}+\tfrac{1}{y+1}+\tfrac{1}{z+1}$ and cancelling denominators (we can do this legally since $x$, $y$, $z>1$ meaning $s < \tfrac 32$), we get $s^2\le 3-2s$, so $s^2+2s-3\le 0$. Factoring, $(s+3)(s-1)\le 0$. As $s$ is clearly positive, we have $s\le 1$, as desired. $\square$

Set $\frac 1{x+1}=a,\frac1{y+1}=b,\frac1{z+1}=c.$ Letting $f(x)=\frac{x^2}{1-2x},$ we can check that $\frac 1{x^2-1}+\frac1{y^2-1}+\frac1{z^2-1}=f(a)+f(b)+f(c).$ Next, from $x,y,z>1$ we get $0<a,b,c< \tfrac12.$ Now we prove a stronger version of the contrapositive: that if $a+b+c>1$ then $f(a)+f(b)+f(c)>1.$ It is easy to check that $f(x)$ is increasing and convex in $(0,\tfrac 12)$ so by Jensen's we have \[f(a)+f(b)+f(c) \ge 3f\left(\frac{a+b+c}3\right)>3f\left(\frac13\right)=1\]which finishes.

ok Let $s = \sum_{\mathrm{cyc}} \frac{1}{x+1}$ Now $$\left( \frac{1}{x^2-1}+ \frac{1}{y^2-1} + \frac{1}{z^2-1} \right) \left( \frac{x-1}{x+1} + \frac{y-1}{y+1} + \frac{z-1}{z+1} \right) \ge s^2$$$$2 \left(\frac{3}{2}-s \right) \ge s^2$$$$0 \ge s^2 +2s-3 $$$$0 \ge (s+3)(s-1)$$$$1 \ge s$$And we are done $\square$

By partial fraction decomposition, we have that $\tfrac{1}{x^2-1} = \tfrac{1}{2(x-1)} - \tfrac{1}{2(x+1)}$. Thus, the given condition becomes $\sum_{cyc} \frac{1}{x-1} = 2 + \sum_{cyc} \frac{1}{x+1}$, and the inequality we are trying to prove is equivalent to \[\sum_{cyc} \frac{1}{x-1} \le 3.\]By Holder, we have \[\left(\sum_{cyc} \frac{x+1}{x-1}\right)\left(\sum_{cyc} \frac{1}{x^2-1}\right) = \left(\sum_{cyc} \frac{x+1}{x-1}\right) \ge \left(\sum_{cyc} \frac{1}{x-1}\right)^2.\]Let $u = \sum_{cyc} \tfrac{1}{x-1}$. Then, the above inequality is equivalent to $3+2u\ge u^2$, which rearranges as $(u+1)(u-3) \le 0$. This implies $u\in [-1,3]$, which finishes. $\blacksquare$

This should also work. Let $a=\frac{1}{x+1}, b=\frac{1}{y+1}\text{ and }c=\frac{1}{z+1}$ so we want to prove that $\sum_{cyc}a\le1$ Now by rearranging we can get $x-1=\frac{1}{a}-2, y-1=\frac{1}{b}-2\text{ and }z-1=\frac{1}{c}-2$ so $\sum_{cyc}\frac{1}{x^2-1}=\sum_{cyc}\frac{1}{(x-1)(x+1)}=\sum_{cyc}a\cdot\frac{1}{\frac{1}{a}-2}=\sum_{cyc}\frac{a^2}{1-2a}$ Furthermore using this yields $$1=\sum_{cyc}\frac{1}{x^2-1}=\sum_{cyc}\frac{a^2}{1-2a}\overset{\text{T2'S}}{\ge}\frac{\left(\sum_{cyc}a\right)^2}{3-2\sum_{cyc}a}$$Thus we have that $3-2\sum_{cyc}a\ge\left(\sum_{cyc}a\right)^2$ is true, however notice that this is equivalent to $0\ge\left(\sum_{cyc}a+3\right)\left(\sum_{cyc}a-1\right)$ which is true if and only if $-3\le\sum_{cyc}a\le1$ However as $\sum_{cyc}a>0$ we must have $\sum_{cyc}a\le1$ which is the same as $\sum_{cyc}\frac{1}{x+1}\le1$ $\blacksquare$.

It suffices to prove that \[\sum_{\text{cyc}} \frac{1}{x+1} \le \sum_{\text{cyc}} \frac{1}{x^2-1}\]\[\Longleftrightarrow \sum_{\text{cyc}} \frac{2-x}{x^2-1} \ge 0\]Let $a = \frac{1}{x^2-1}$ and similar for $b$ and $c$ such that \[a+b+c = \frac{1}{x^2-1} + \frac{1}{y^2-1} + \frac{1}{z^2-1} = 1\]By substituting $x = \sqrt{\frac{1}{a}+1}$, it suffices to prove \[\sum_{\text{cyc}} 2a - \sqrt{a^2+a} \ge 0\]\[\Longleftrightarrow \sum_{\text{cyc}} \sqrt{a^2+a} \le 2\]Let $f(x) = \sqrt{x^2+x}$. We have that \[f''(x) = -\frac{1}{4(x^2+x)^{\frac{3}{2}}} < 0\]for all $x$. Thus by Jensen's with $a+b+c = 1$ we have \[\sum_{\text{cyc}} \sqrt{a^2+a} \le 3 \sqrt{\left(\frac{1}{3}\right)^2+\left(\frac{1}{3}\right)} = 2\]as desired.

Let $a=\frac{1}{x^2-1}, b=\frac{1}{y^2-1}, c=\frac{1}{z^2-1}$. Then we have $a+b+c=1$ and we want to prove that: $$\sum{\frac{1}{\sqrt{\frac{1}{a}+1}+1}} \leq 1$$Now calculating first derivative and doing the tangent line trick we want to prove that: $${\frac{1}{\sqrt{\frac{1}{x}+1}+1}} \leq \frac{1}{4} x+\frac{1}{4} \overset{\text{A bit of transformation}}{\Leftrightarrow} 0\leq (3x-1)^2$$Summing this inequality for $a,b,c$ we get: $$\sum{\frac{1}{\sqrt{\frac{1}{a}+1}+1}} \leq \frac{1}{4}(\sum{a})+\frac{3}{4}=1$$

Let $$a = \frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}.$$By Cauchy, we know that $$(\frac{x-1}{x+1}+\frac{y-1}{y+1}+\frac{z-1}{z+1})\cdot(\frac{1}{x^2-1}+\frac{1}{y^2-1}+\frac{1}{z^2-1})$$$$\ge (\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1})^2.$$Since $$\frac{x-1}{x+1}+\frac{y-1}{y+1}+\frac{z-1}{z+1} = 3-2a,$$$3-2a\ge a^2$ and $a\le 1$.

Let $a=\frac{1}{x^2-1}$ and simialrly for other variables. So $a+b+c=1$ and we wish to prove that $$\sum \frac{1}{\sqrt{\frac{1}{a}+1}+1} \leq 1$$which follows from Jensens as the second dervative is negative.

omg I can solve inequalities no way

XOOOOOOOOOOOOOOOÕŌØÔÖÒÓ WE CAN : XOOK $x$ UNTIL THE INEQUALITY IS SKIB. THEN WE WANT TO SHOW \[\sum_{\text{cyc}}\left(\frac{1}{x-1}-\frac{1}{x+1}\right)\ge 2\Longleftrightarrow \sum_{\text{cyc}}\frac{1}{x-1}\ge 3.\]IF : $t=\frac{1}{x+1}$ THEN \[\frac{1}{x-1}=\frac{1}{\frac{1}{t}-2}=\frac{t}{1-2t}=\frac{1/2}{1-2t}-\frac12\;\;.\]OOK! NOW WE WANT \[\sum_{\text{cyc}}\frac{1}{1-2t}\ge 9\]BUT THATS JUST CAUCHY!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Let $x,y,z \ge 1$ and $\frac{1}{x^2-1} + \frac{6}{y^2-1} + \frac{1}{z^2-1} = 1. $ Prove that$$\frac{1}{x+1} + \frac{6}{y+1} + \frac{1}{z+1} \le 2$$