cyclic sum 1 / (a(b+1)) + ... >= 3 / (1+abc)

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Arne

3660 posts

Arne

#1 h Aug 17, 2003, 8:56 AM • 11 Y

Y by Adventure10, Matherer9654, Mango247, EntropiaAwake, EntropiaAwake, MS_asdfgzxcvb, and 5 other users

Let $a$ , $b$ , $c$ be positive real numbers. Prove the inequality
$\frac{1}{a\left(b+1\right)}+\frac{1}{b\left(c+1\right)}+\frac{1}{c\left(a+1\right)}\geq \frac{3}{1+abc}.$

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darij grinberg

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darij grinberg

#2 h Apr 6, 2004, 5:32 PM • 17 Y

Y by shinichiman, HYP135peppers, MathArt4, Hermitianism, ThisIsASentence, AlastorMoody, Adventure10, Matherer9654, Mango247, MS_asdfgzxcvb, and 7 other users

The same inequality appeared in a Kolmogorov cup (a Russian "math fight" competition) in the following form:

Problem. For any three positive numbers a, b, c, prove the inequality

$\left( 1+abc\right) \left(\frac{1}{a\left( 1+b\right) }+\frac{1}{b\left( 1+c\right) }+\frac{1}{c\left( 1+a\right) }\right) \geq 3$ .

___________________________________________________________________________

I think this is a beautiful inequality. But, as humble as I always am

, I think I also have a very beautiful proof of it. I really can't resist posting it here (you can try searching for other proofs, since I really don't think this was the intended solution!).

In fact, in general we have $abc \neq 1$ , we cannot apply Harazi's substitution $a=\frac{v}{u}$ , $b=\frac{w}{v}$ , $c=\frac{u}{w}$ with positive u, v, w. But we can substitute $a=k\cdot \frac{v}{u}$ , $b=k\cdot \frac{w}{v}$ , $c=k\cdot \frac{u}{w}$ , where k, u, v, w are positive real numbers. Then $abc=k^{3}$ , and our inequality simplifies to

$\left( 1+k^{3}\right) \left(\frac{u}{k\left( v+kw\right) }+\frac{v}{k\left( w+ku\right) }+\frac{w}{k\left( u+kv\right) }\right) \geq 3$ .

This can be rewritten as

$\frac{1+k^{3}}{k}\cdot \left(\frac{u}{v+kw}+\frac{v}{w+ku}+\frac{w}{u+kv}\right) \geq 3$ .

Now this is a generalization of Nesbitt's inequality. For proving it, we define A = v + kw, B = w + ku, C = u + kv, and then we can readily find

$u = \frac{C+k^{2}B-kA}{1+k^{3}}$ ;
$v = \frac{A+k^{2}C-kB}{1+k^{3}}$ ;
$w = \frac{B+k^{2}A-kC}{1+k^{3}}$ .

Thus, the inequality we are intending to prove can be rewritten as

$\frac{1+k^{3}}{k}\cdot \left( \frac{C+k^{2}B-kA}{A\left( 1+k^{3}\right) }+\frac{A+k^{2}C-kB}{B\left( 1+k^{3}\right) }+\frac{B+k^{2}A-kC}{C\left( 1+k^{3}\right) }\right) \geq 3$ .

In other words,

$\frac{1}{k}\cdot \left( \frac{C+k^{2}B-kA}{A}+\frac{A+k^{2}C-kB}{B}+\frac{B+k^{2}A-kC}{C}\right) \geq 3$ .

Multiply with k and get

$\frac{C+k^{2}B-kA}{A}+\frac{A+k^{2}C-kB}{B}+\frac{B+k^{2}A-kC}{C}\geq 3k$ .

This can be rearranged (not in the meaning of using the rearrangement inequality) into

$\left( \frac{B}{C}+k^{2}\frac{C}{B}\right)+\left( \frac{C}{A}+k^{2}\frac{A}{C}\right)+\left( \frac{A}{B}+k^{2}\frac{B}{A}\right)-3k \geq 3k$ .

In other words,

$\left( \frac{B}{C}+k^{2}\frac{C}{B}\right)+\left( \frac{C}{A}+k^{2}\frac{A}{C}\right)+\left( \frac{A}{B}+k^{2}\frac{B}{A}\right) \geq 6k$ .

But this is trivial since AM-GM gives $\frac{x}{y}+k^{2}\frac{y}{x}\geq 2k$ .

___________________________________________________________________________

EDIT: See also

http://www.mathlinks.ro/Forum/viewtopic.php?t=48261
http://www.mathlinks.ro/Forum/viewtopic.php?t=33550
http://www.mathlinks.ro/Forum/viewtopic.php?t=19469 problem 2

for this and related inequalities.

This inequality seems to have appeared in 1000 different places. You can discuss about its history at http://www.mathlinks.ro/Forum/viewtopic.php?t=85696 . As for now, the following sources of the above inequality are known:

- Balkan MO 2006, Problem 1, proposed by Sotiris Louridas from Greece;
- Crux Mathematicorum 1998, proposed by Mohammed Aassila;
- (maybe) the book "Old and New Inequalities" (at least the stronger version, http://www.mathlinks.ro/Forum/viewtopic.php?t=6044 , appeared in this book);
- UK Correspondence Course 2005;
- a Greek book on inequalities written by Skombris and Babis Stergiou (this is the MathLinks user Stergiu), where "Kömal 1993" is given as a source for this inequality (but this source may be imprecise);
- proposed by the Russian mathematician D. P. Mavlo in the Bulgarian journal Mathematica (number 4, 1987);
- Kvant problem M1136, number 11-12, 1988;
- USA IMO team training 1993;
- South African IMO team training;
- Titu Andreescu's book "Mathematical Miniatures";
- 6th Kolmogorov Cup, 1-8 December 2002, 1 round, 1 league;
- Indian IMO selection test 1997;
- Armenian National Olympiad 1996;
- Kiran Kedlaya's inequality notes (?);
- Po-Shen Loh's Inequality Notes (see http://www.mathlinks.ro/Forum/viewtopic.php?t=19469 ).

It is a very interesting inequality, but I hope we won't see it on a future olympiad anymore.

Darij

This post has been edited 5 times. Last edited by darij grinberg, Aug 2, 2007, 1:42 PM

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fuzzylogic

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fuzzylogic

#3 h Aug 5, 2004, 9:41 PM • 11 Y

Y by TheOneYouWant, vsathiam, xyzz, ZHEKSHEN, Adventure10, Mango247, MS_asdfgzxcvb, and 4 other users

darij grinberg wrote:

$\displaystyle \frac{1+k^3}{k} \cdot \left(\frac{u}{v+kw}+\frac{v}{w+ku}+\frac{w}{u+kv}\right) \geq 3$ .

Indeed, a beautiful problem and beautiful substitution. But I am surprised you didn't finish it up using your favorite Cauchy-Schwarz in Engel form:

$\displaystyle \frac{1+k^3}{k} \cdot \left(\frac{u}{v+kw}+\frac{v}{w+ku}+\frac{w}{u+kv}\right) \geq \frac{1+k^3}{k(k+1)} \cdot \frac{(u+v+w)^2}{uv+vw+wu}$ .

The first factor is $\geq 1$ since $1+k^3\geq k^2+k$ by rearrangement.

The second factor is $\geq 3$ which is well-known.

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fuzzylogic

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fuzzylogic

#4 h Aug 6, 2004, 1:35 AM • 8 Y

Y by mmberrymm, Adventure10, k_rs52, ehuseyinyigit, Mango247, and 3 other users

I tried brute-force, it isn't very hard. Just grouping properly and then apply AM-GM.

But harazi has a beautiful proof for a stronger version at:

http://www.mathlinks.ro/Forum/viewtopic.php?t=6044

He showed that:

$\displaystyle \left( 1+abc\right) \left(\frac{1}{a\left( 1+b\right) }+\frac{1}{b\left( 1+c\right) }+\frac{1}{c\left( 1+a\right) }\right) + 3$

$\displaystyle = \frac{1+abc+a+ab}{a+ab}+\frac{1+abc+b+bc}{b+bc}+\frac{1+abc+c+ca}{c+ca}$

$\displaystyle = \frac{1+a}{ab+a}+\frac{b+1}{bc+b}+\frac{c+1}{ca+c} + \frac{b(c+1)}{b+1}+\frac{c(a+1)}{c+1}+\frac{a(b+1)}{a+1}$

$\displaystyle \geq\frac{3}{\sqrt[3]{abc}} + 3\sqrt[3]{abc}$ which is of course $\geq 6$ .

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isotomion

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isotomion

#5 h Aug 6, 2004, 11:34 AM • 3 Y

Y by Adventure10, Mango247, and 1 other user

Thanks indeed, both your and Harazi's proofs are much better than mine. Well, but to my defence I must say that what I did - with the last substitution - was essentially equivalent to proving Cauchy-Schwarz in the Engel form. And both your (Fuzzylogic)'s and my proofs can be modified to prove the stronger inequality from http://www.mathlinks.ro/Forum/viewtopic.php?t=6044 !

Darij

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math_sipo

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math_sipo

#6 h Aug 7, 2004, 9:53 PM • 3 Y

Y by Adventure10, Mango247, and 1 other user

you can just be courageous and try the traditional way it's very easy man!

(1+abc)(bc+ac+ab+3abc+bc+ca+ab)>=3(abc+abc(a+b+c)+abc(ab+bc+ac)+abc)
just try it it's easy !
bye

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blahblahblah

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blahblahblah

#7 h Nov 9, 2004, 12:02 PM • 5 Y

Y by Adventure10, Mango247, and 3 other users

I cleared denominators and hacked it out using AM-GM. I am horrible at fractional inequalities like that, but the whole thing comes down to:

$\displaystyle \left(\sum_{cyc} ab^2 +\sum_{cyc} a^3b^2c^2\right) +\left(\sum_{cyc} a^3bc^2 +\sum_{cyc} ab \right)-2\sum_{sym} a^2b^2c-2\sum_{sym} a^2bc\geq 0$ , and the terms in brackets are easily seen to be bigger than the terms with negative signs by AM-GM

This post has been edited 1 time. Last edited by blahblahblah, Nov 11, 2004, 3:44 AM

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nttu

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nttu

#8 h Nov 9, 2004, 4:48 PM • 6 Y

Y by TheOneYouWant, Adventure10, ehuseyinyigit, Mango247, and 2 other users

The inequality is equivalent to
$\sum_{cyc}\frac{abc+1}{a(1+b)}\geq 3$
$\sum_{cyc}(\frac{1+a}{a(1+b)}+\frac{abc+ab}{a(1+b)}) \geq 6$
This ineq is right .( using AM_GM 3 times)

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Soarer

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Soarer

#9 h Nov 9, 2004, 4:58 PM • 3 Y

Y by Adventure10, Mango247, and 1 other user

hi nttu, your solution is very nice!!
Anyway, may i know how can you think of converting (abc+1)/a(1+b) to (abc+ab)/a(1+b)+(1+a)/a(1+b) - 1 ??
Thanks!

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Cezar Lupu

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Cezar Lupu

#10 h Aug 29, 2005, 10:27 PM • 7 Y

Y by Understandingmathematics, vanwij, Adventure10, ladyMS, Mango247, and 2 other users

Nice and easy. In fact, this is too easy!

We have
$\frac{1+abc}{a+ab}=\frac{1+a+ab+abc}{a+ab}-1=\frac{1+a}{a(1+b)}+\frac{b(1+c)}{1+b}-1$ .
Hence, rewrite the inequality in the form:

$\frac{1+a}{a(1+b)}+\frac{b(1+c)}{1+b}+\frac{1+b}{b(1+c)}+\frac{c(1+a)}{1+c}+\frac{1+c}{c(1+a)}+\frac{a(1+b)}{1+a}\geq 6$ . This one follows immediately from AM-GM.

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Cezar Lupu

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Cezar Lupu

#11 h Aug 31, 2005, 7:49 PM • 4 Y

Y by Adventure10, Mango247, and 2 other users

Actually, this inequality can be proved using the following inequality of D.M.Pavlo which states that for any $a,b,c>0$ the following inequality holds: $3+a+b+c+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\geq 3\cdot\frac{(a+1)(b+1)(c+1)}{1+abc}$ . So we only need to prove that $\frac{1}{a+ab}+\frac{1}{b+bc}+\frac{1}{c+ca}\geq\frac{1}{{(a+1)(b+1)(c+1)}\cdot(3+a+b+c+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{a}{b}+\frac{b}{c}+\frac{c}{a})}$ . Well, by making some calculaions here, we get $\sum\frac{a+1)(b+1)}{c}\geq 3+\sum a+\sum\frac{1}{a}+\sum\frac{a}{b}$ which is equivalent with the following inequality:

$\sum\frac{ab}{c}+\sum\frac{a+b}{c}+\frac{1}{a}\geq 3+\sum a+\sum\frac{1}{a}+\sum\frac{a}{b}$ so the inequality reduces finally to this one: $\sum\frac{ab}{c}+\sum\frac{a}{c}\geq 3+\sum a$ but from here it is easy, since we do know that $\frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b}\geq a+b+c$ and $\frac{a}{c}+\frac{b}{a}+\frac{c}{b}\geq 3$ .Both of then follow from AM-GM.

Hope I haven't made any mistakes

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iandrei2

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iandrei2

#12 h Sep 2, 2005, 12:56 PM • 2 Y

Y by Adventure10 and 1 other user

Hello,

I'll post my attempt. It is based on computations, but it took me about 15 minutes to solve it this way.

Attachments:

ineg.pdf (40kb)

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silouan

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silouan

#13 h Dec 16, 2005, 4:31 PM • 4 Y

Y by vsathiam, Adventure10, Mango247, and 1 other user

I will also post a solution.
After clearing the denominators the ineq is equivalent to

$ab(b+1)(ca-1)^{2}+bc(c+1)(ab-1)^{2}+ca(a+1)(bc-1)^{2}\geq 0$ which is true

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arqady

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arqady

#14 h Apr 21, 2006, 3:53 AM • 5 Y

Y by LeMagnifiqueEp, Adventure10, Mango247, and 2 other users

X-man wrote:

If $a, b, c \in \mathbb R$ are positive then prove $\frac{1}{a(b+1)} + \frac{1}{b(c+1)} + \frac{1}{c(a+1)} \ge \frac{3}{1 + abc}$

$\frac{1}{a(b+1)} + \frac{1}{b(c+1)} + \frac{1}{c(a+1)} \ge \frac{3}{1 + abc}\Leftrightarrow\sum_{cyc}(bc^2+bc)(ab-1)^2\geq0.$

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arpit goel

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arpit goel

#15 h Apr 29, 2006, 7:22 PM • 4 Y

Y by Adventure10, Adventure10, Mango247, and 1 other user

hi

i hv done the problem

clear the denominator and solve
in the end u wud be left with am gm inequality application

the process is quite long but not very tough to think

wud like to see a short proof

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sqing

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sqing

#49 h Mar 30, 2019, 3:55 PM • 2 Y

Y by Adventure10, Mango247

Arne wrote:

Let $a$ , $b$ , $c$ be positive real numbers. Prove the inequality
$\frac{1}{a\left(b+1\right)}+\frac{1}{b\left(c+1\right)}+\frac{1}{c\left(a+1\right)}\geq \frac{3}{1+abc}.$

Let $a,b,c$ be positive real numbers . Prove that $\frac{1}{ab(b+1)(c+1)}+\frac{1}{bc(c+1)(a+1)}+\frac{1}{ca(a+1)(b+1)}\geq\frac{3}{(1+abc)^2}.$

sqing wrote:

Let $a_1,a_2,\cdots,a_n\geq 1$ $(n\ge 3)$ . Prove that $\frac{1}{a_1\left(a_2+1\right)}+\frac{1}{a_2\left(a_3+1\right)}+\cdots+\frac{1}{a_{n-1}\left(a_n+1\right)}+\frac{1}{a_n\left(a_1+1\right)}\geq \frac{n}{1+a_1a_2\cdots a_n}.$

Let $a_1,a_2,\cdots,a_n\geq 1$ $(n\ge 3)$ . Prove or disprove $\frac{1}{a_1a_2\left(a_2+1\right)\left(a_3+1\right)}+\frac{1}{a_2a_3\left(a_3+1\right)\left(a_4+1\right)}+\cdots+\frac{1}{a_{n-1}a_n\left(a_n+1\right)\left(a_1+1\right)}+\frac{1}{a_na_1\left(a_1+1\right)\left(a_2+1\right)}\geq \frac{n}{(1+a_1a_2\cdots a_n)^2}.$

This post has been edited 2 times. Last edited by sqing, Mar 30, 2019, 4:03 PM

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mihaig

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mihaig

#50 h Mar 30, 2019, 4:32 PM • 1 Y

Y by Adventure10

sqing wrote:

Arne wrote:

Let $a$ , $b$ , $c$ be positive real numbers. Prove the inequality
$\frac{1}{a\left(b+1\right)}+\frac{1}{b\left(c+1\right)}+\frac{1}{c\left(a+1\right)}\geq \frac{3}{1+abc}.$

Let $a,b,c$ be positive real numbers . Prove that $\frac{1}{ab(b+1)(c+1)}+\frac{1}{bc(c+1)(a+1)}+\frac{1}{ca(a+1)(b+1)}\geq\frac{3}{(1+abc)^2}.$

Let's notice that my inequality refines the Crux problem used in 2006.

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ytChen

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ytChen

#51 h Sep 19, 2019, 2:07 AM • 2 Y

Y by Adventure10, Mango247

sqing wrote:

Let $a_1,a_2,\cdots,a_n\geq 1$ $(n\ge 3)$ . Prove that $\frac{1}{a_1\left(a_2+1\right)}+\frac{1}{a_2\left(a_3+1\right)}+\cdots+\frac{1}{a_{n-1}\left(a_n+1\right)}+\frac{1}{a_n\left(a_1+1\right)}\geq \frac{n}{1+a_1a_2\cdots a_n}.$

Solution. Stipulate $a_{n+1}=a_1$ and $a_{n+2}=a_2$ .
$\begin{align*}\displaystyle\sum_{i=1}^n\left(\frac{1+a_1a_2\cdots a_n}{a_i\left(a_{i+1}+1\right)}+1\right)=&\displaystyle\sum_{i=1}^n\frac{1+a_i+a_ia_{i+1}(1+a_1\cdots a_{i-1}a_{i+2}\cdots a_n)}{a_i\left(a_{i+1}+1\right)}\\ =&\displaystyle\sum_{i=1}^n\left(\frac{1+a_i}{a_i\left(a_{i+1}+1\right)}+\frac{a_{i+1}(1+a_1\cdots a_{i-1}a_{i+2}\cdots a_n)}{a_{i+1}+1}\right)\\ \left(\because a_i\ge1\,\forall i\right)\quad\ge&\displaystyle\sum_{i=1}^n\left(\frac{1+a_i}{a_i\left(a_{i+1}+1\right)}+\frac{a_{i+1}(1+a_{i+2})}{a_{i+1}+1}\right)\\ =&\displaystyle\sum_{i=1}^n\frac{1+a_i}{a_i\left(a_{i+1}+1\right)}+\sum_{i=1}^n\frac{a_{i+1}(1+a_{i+2})}{a_{i+1}+1}\\ \left(\text{by AM-GM}\right)\quad\ge&\frac{n}{\sqrt[n]{a_1a_2\cdots a_n}}+n\sqrt[n]{a_1a_2\cdots a_n}\ge2n. \end{align*}$ Therefore we have $\displaystyle\sum_{i=1}^n\frac{1+a_1a_2\cdots a_n}{a_i\left(a_{i+1}+1\right)}\ge n$ thereby $\displaystyle\sum_{i=1}^n\frac{1}{a_i\left(a_{i+1}+1\right)}\ge\frac{n}{1+a_1a_2\cdots a_n}$ . As required. $\blacksquare$

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xzlbq

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xzlbq

#53 h Sep 19, 2019, 4:00 PM • 1 Y

Y by Adventure10

Arne wrote:

Let $a$ , $b$ , $c$ be positive real numbers. Prove the inequality
$\frac{1}{a\left(b+1\right)}+\frac{1}{b\left(c+1\right)}+\frac{1}{c\left(a+1\right)}\geq \frac{3}{1+abc}.$

stronger is this:

Let $a,b,c>0$ ,prove that

$\left( {\frac {1}{a \left( b+1 \right) }}+{\frac {1}{b \left( c+1 \right) }}+{\frac {1}{c \left( 1+a \right) }} \right) \left( 1+abc \right) \geq {\frac {27}{8}}\,{\frac { \left( b+c \right) \left( c+a \right) \left( a+b \right) }{ \left( a+b+c \right) \left( ab+bc+ac \right) }}$

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sqing

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sqing

#54 h Dec 7, 2019, 12:46 PM • 2 Y

Y by Adventure10, Mango247

sqing wrote:

Let $a_1,a_2,\cdots,a_n\geq 1$ $(n\ge 3)$ . Prove that $\frac{1}{a_1\left(a_2+1\right)}+\frac{1}{a_2\left(a_3+1\right)}+\cdots+\frac{1}{a_{n-1}\left(a_n+1\right)}+\frac{1}{a_n\left(a_1+1\right)}\geq \frac{n}{1+a_1a_2\cdots a_n}.$

Let $a_1,a_2,\cdots,a_n\geq 1$ $(n\ge 3)$ . Prove that $\frac{1}{a_1\left(a_2+n-2\right)}+\frac{1}{a_2\left(a_3+n-2\right)}+\cdots+\frac{1}{a_{n-1}\left(a_n+n-2\right)}+\frac{1}{a_n\left(a_1+n-2\right)}\geq \frac{n}{n-2+a_1a_2\cdots a_n}.$

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sqing

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sqing

#55 h Mar 24, 2021, 1:43 PM

Y by

Let $a_1,a_2,\cdots,a_n>0$ $(n\ge 2)$ . Prove that $\frac{1}{a_1\left(a_1+n-2\right)}+\frac{1}{a_2\left(a_2+n-2\right)}+\cdots+\frac{1}{a_n\left(a_n+n-2\right)}\geq \frac{n}{n-2+a_1a_2\cdots a_n}.$ (Lijvzhi)

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JustKeepRunning

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JustKeepRunning

#57 h Jul 29, 2021, 7:13 PM

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A pretty hard problem (idk why, but solutions like those in @nttu's post are almost impossible to find)

Suppose that $abc=k^3$ . Then we can substitute $a=\frac x y k, b = \frac z x k, c = \frac y z k$ . (This is a critical step. Notice how we did not apply the typical cyclic substitution. As in @dg's post, this simplifies down to showing that $\frac{k^3+1}{k}\sum_{cyc} \frac{x}{ky+z}\geq 3$ . Of course, we forces a square into the numerator to get an equivalent inequality of $\frac{1+k^3}{k}\sum_{cyc}\frac{x^2}{kxy+xz}\geq 3$ . Now we have from Engel that $\text{LHS}\geq \frac{1+k^3} k \cdot \frac{(x+y+z)^2}{(k+1)(xy+yz+zx)} = \frac{1+k^3}{k(k+1)}\frac{(x+y+z)^2}{xy+yz+zx}$ . Notice that $k^3+1\geq k^2+k\leftrightarrow (k-1)^2(k+1)\geq 0,$ so we have that the first fraction $\geq 1,$ and it is well known that the second fraction $\geq 3,$ so we are done.

In @fuzzylogic's sol, he says that $1+k^3\geq k^2+k$ follows from rearrangement. How can this be possible? I thought rearrangements was only for products?

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silouan

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silouan

#58 h Jul 30, 2021, 1:14 PM

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@above
Consider the tuples $(k^2,k)$ and $(k,1)$ . They are equally sorted, so the result follows indeed by the rearrangement inequality.

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ehuseyinyigit

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ehuseyinyigit

#59 h Oct 8, 2023, 5:35 PM

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Came here to generalize but Mr.Sqing had done the whole work

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ehuseyinyigit

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ehuseyinyigit

#60 h Nov 12, 2023, 6:22 PM

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Finally I have made a generalization to this problem.

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ehuseyinyigit

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ehuseyinyigit

#61 h Nov 12, 2023, 6:23 PM

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The original problem is a special case of the following generalization where
$n=1$ .

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ehuseyinyigit

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ehuseyinyigit

#62 h Nov 12, 2023, 6:24 PM

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Generalization 1
Let $a_{1},a_{2},\cdots,a_{3n}$ be positive reals ( $n\geq 1$ ). Then prove that

$\sum_{cyc}{\left(\dfrac{1}{a_{1}a_{2}\cdots a_{n}\left(a_{n+1}a_{n+2}\cdots a_{2n}+1\right)}\right)}\geq \dfrac{3n}{\prod{a_{1}}+1}$

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sqing

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sqing

#63 h Oct 17, 2024, 9:51 AM

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Let $a$ , $b$ , $c$ be positive real numbers. Prove the inequality
$\frac{a}{bc(1+b)}+\frac {b}{ca(1+c)}+\frac {c}{ab(1+a)}\geq \frac {3}{ 1+abc}$

Arne wrote:

Let $a$ , $b$ , $c$ be positive real numbers. Prove the inequality
$\frac{1}{a\left(b+1\right)}+\frac{1}{b\left(c+1\right)}+\frac{1}{c\left(a+1\right)}\geq \frac{3}{1+abc}.$

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This post has been edited 1 time. Last edited by sqing, Mar 13, 2025, 2:21 PM

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