Let $x,y,z \ge 1$ be real numbers such that \[ \frac{1}{x^2-1} + \frac{1}{y^2-1} + \frac{1}{z^2-1} = 1. \] Prove that \[ \frac{1}{x+1} + \frac{1}{y+1} + \frac{1}{z+1} \le 1. \]
Problem
Source: ELMO 2003 Problem 4
Tags: inequalities, Hi
29.12.2012 02:55
29.12.2012 19:23
29.12.2012 19:25
A less tricky way for less trickier people.
10.11.2017 04:29
11.11.2017 10:22
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*}
11.11.2017 13:42
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.
18.03.2018 06:15
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.
18.03.2018 15:23
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
16.11.2019 15:27
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 $s \leq 1$
16.11.2019 15:57
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 $a=b=c=\frac{1}{3}$.
16.11.2019 19:42
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. $\Box$
17.10.2020 22:28
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.
31.12.2020 07:23
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. $\square$
31.12.2020 08:33
From where do you get this? \[3-2A \ge A^2\]
31.12.2020 08:43
\[\left(\sum \frac{1}{x^2-1}\right)\left(\sum \frac{x-1}{x+1}\right) = \sum \left( 1 - \frac{2}{x+1} \right) = 3 - 2A \]
31.12.2020 08:51
Evan Chen made a typo. It is Po-Shen-Loh
31.12.2020 08:59
Um, no, he didn't. Po-Ru Loh is a real person and I believe won 1999 Countdown round in MC Nats.
31.12.2020 14:13
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. $\square$
31.12.2020 14:41
By AM-GM, $\frac{1}{x^2-1}+\frac{x-1}{x+1} \ge \frac{2}{x+1}$. $\sum (\frac{1}{x^2-1}+(1- \frac{2}{x+1})) \ge \sum \frac{2}{x+1} \Longrightarrow 1 \ge \sum \frac{1}{x+1}$
22.11.2023 07:54
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 $\blacksquare$
24.11.2023 18:06
01.12.2023 09:02
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.
10.12.2023 04:24
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. $\square$
04.01.2024 17:46
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.
21.02.2024 03:18
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 $\square$
13.06.2024 00:14
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. $\blacksquare$
20.06.2024 18:44
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 $\sum_{cyc}\frac{1}{x+1}\le1$ $\blacksquare$.
07.08.2024 10:04
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.
28.08.2024 00:14
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: $$\sum{\frac{1}{\sqrt{\frac{1}{a}+1}+1}} \leq \frac{1}{4}(\sum{a})+\frac{3}{4}=1$$
03.11.2024 04:33
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 $a\le 1$.
29.11.2024 22:39
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.
11.01.2025 04:06
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!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
12.01.2025 11:08
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$$\frac{1}{x+1} + \frac{6}{y+1} + \frac{1}{z+1} \le 2$$