USA 97 [1/(b^3+c^3+abc) + ... >= 1/(abc)]

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Maverick

331 posts

Maverick

#1 h Sep 12, 2003, 3:57 PM • 6 Y

Y by thuanz123, Adventure10, HWenslawski, Mango247, Blue_banana4, ItsBesi

Prove that, for all positive real numbers $a$ , $b$ , $c$ , the inequality
$\frac {1}{a^3 + b^3 + abc} + \frac {1}{b^3 + c^3 + abc} + \frac {1}{c^3 + a^3 + abc} \leq \frac {1}{abc}$
holds.

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Arne

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Arne

#2 h Sep 12, 2003, 4:04 PM • 6 Y

Y by Akababa, Adventure10, HWenslawski, Tafi_ak, jianglianjun, Mango247

We have x³ + y³ \geq x²y + xy² for all x, y > 0 by rearrangement. So
\sum 1/(a³ + b³ + abc) \leq \sum 1/(a²b + ab² + abc) = 1/abc by simple computation.

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blahblahblah

4012 posts

blahblahblah

#3 h Mar 13, 2005, 11:38 PM • 5 Y

Y by Binomial-theorem, Akababa, Adventure10, Mathmick51, Mango247

$a^3+b^3\geq ab^2+a^2b$ , so we get that:

$\sum \frac{abc}{a^3+b^3+abc}\leq \sum\frac{abc}{ab(a+b+c)} = \sum\frac{c}{a+b+c}=1$

This post has been edited 1 time. Last edited by blahblahblah, Jan 9, 2006, 6:17 AM

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paladin8

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paladin8

#4 h Mar 14, 2005, 1:24 AM • 2 Y

Y by Adventure10, Mango247

Ok, thanks.

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blahblahblah

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blahblahblah

#5 h Mar 14, 2005, 1:51 AM • 2 Y

Y by Adventure10, Mango247

Out of curiosity, why would you even want to let $abc=1$ ? You can apply Jensen's in the exact same fashion without de-homogenizing.

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paladin8

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paladin8

#6 h Mar 14, 2005, 2:03 AM • 2 Y

Y by Adventure10, Mango247

That's a good point. I don't know why; I was just working with it and decided to set $abc=1$ . Then I saw the solution with Jensen's after. So it was merely a detour in my thought process. Thanks for pointing that out. It's probably a clearer solution that way.

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MysticTerminator

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MysticTerminator

#7 h Jul 25, 2005, 3:27 AM • 2 Y

Y by Adventure10, Mango247

this has been discussed a couple times

I want to say it's USAMO 97, not sure. the idea is to say a^3 + b^3 >= ab(a+b) to make each denominator smaller.

[Moderator edit: For this problem, see also
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=58140
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=19469
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=69177
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=55677
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=49889
]

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splintercell990

43 posts

splintercell990

#8 h Jan 15, 2006, 3:34 AM • 3 Y

Y by Binomial-theorem, Adventure10, Mango247

I came across this thread looking for practice USAMO inequalities so I apologize for reviving such an old thread, but here's a slightly different solution using symmetry and bunching:

Click to reveal hidden text

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AdithyaBhaskar

652 posts

AdithyaBhaskar

#9 h Mar 18, 2016, 3:04 AM • 1 Y

Y by Adventure10

Some motivation. The summands are just so 'nice'. This makes me feel the urge to separate right hand side into $\frac{a}{abc(a+b+c)},$ etc.
Solution. And miraculously it works. Of course by muirhead and $[3,0] \succ [2,1]$ one has $a^3 + b^3 \geq a^2b + ab^2.$ Then, the left hand side is at most $\sum_{cyc} \frac{a}{abc(a+b+c)} = \frac{1}{abc}.$

This post has been edited 1 time. Last edited by AdithyaBhaskar, Mar 18, 2016, 3:04 AM

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AMN300

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AMN300

#10 h May 16, 2016, 6:08 AM • 2 Y

Y by Adventure10, Mango247

Does this count as a one liner

?
solution

This post has been edited 1 time. Last edited by AMN300, May 17, 2016, 5:48 PM
Reason: asdf

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dovakin123

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dovakin123

#11 h Apr 21, 2017, 11:22 AM • 1 Y

Y by Adventure10

Well I will add another one using Cauchy- Schwarz Inequality

We have using that $\dfrac{a^2}{bc}+\dfrac{b^2}{ca}\ge \dfrac{(a+b)^2}{c(a+b)}\Rightarrow \dfrac{a^2}{bc}+\dfrac{b^2}{ca}+1\ge \dfrac{a+b+c}{c}$ $\Rightarrow \dfrac{1}{\dfrac{a^2}{bc}+\dfrac{b^2}{ca}+1}\le \dfrac{c}{a+b+c}\Rightarrow \sum_{cyc}\dfrac{1}{\dfrac{a^2}{bc}+\dfrac{b^2}{ca}+1}\le 1$

Equality holds iff $a=b=c$ .

This post has been edited 2 times. Last edited by dovakin123, Apr 21, 2017, 11:24 AM

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xzlbq

15849 posts

xzlbq

#12 h Apr 21, 2017, 1:26 PM • 2 Y

Y by Adventure10, Mango247

Maverick wrote:

Prove that, for all positive real numbers $a$ , $b$ , $c$ , the inequality
$\frac {1}{a^3 + b^3 + abc} + \frac {1}{b^3 + c^3 + abc} + \frac {1}{c^3 + a^3 + abc} \leq \frac {1}{abc}$ holds.

Prove that, for all positive real numbers $a$ , $b$ , $c$ , the inequality
$\frac {1}{a^3 + b^3 + abc} + \frac {1}{b^3 + c^3 + abc} + \frac {1}{c^3 + a^3 + abc} \geq \frac {5}{2abc+a^3+b^3+c^3}$ holds.

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xzlbq

15849 posts

xzlbq

#13 h Apr 21, 2017, 2:27 PM • 1 Y

Y by Adventure10

xzlbq wrote:

Prove that, for all positive real numbers $a$ , $b$ , $c$ , the inequality
$\frac {1}{a^3 + b^3 + abc} + \frac {1}{b^3 + c^3 + abc} + \frac {1}{c^3 + a^3 + abc} \geq \frac {5}{2abc+a^3+b^3+c^3}$ holds.

<=>
${\frac {27\,{u}^{6}-54\,{v}^{2}{u}^{4}+10\,{w}^{3}{u}^{3}+27\,{v}^{4}{ u}^{2}-11\,{v}^{2}{w}^{3}u+{v}^{6}+{w}^{6}}{27\,{w}^{3}{u}^{6}-81\,{v} ^{2}{w}^{3}{u}^{4}+27\,{v}^{6}{u}^{3}+11\,{w}^{6}{u}^{3}+54\,{v}^{4}{w }^{3}{u}^{2}-15\,{v}^{2}{w}^{6}u-27\,{v}^{8}u+{w}^{9}+4\,{v}^{6}{w}^{3 }}}\geq {\frac {5}{27\,{u}^{3}-27\,u{v}^{2}+5\,{w}^{3}}}$
<=>

$27\,{w}^{3} \left( 22\,{w}^{3}{u}^{3}-7\,{v}^{2}{w}^{3}u+270\,{u}^{6}- 432\,{v}^{2}{u}^{4}-15\,{v}^{6}+162\,{v}^{4}{u}^{2} \right) +729\,u \left( u-v \right) \left( u+v \right) \left( 3\,{u}^{2}-4\,{v}^{2} \right) \left( -{v}^{2}+3\,{u}^{2} \right) ^{2}\geq 0$
Need to prove

$22\,{w}^{3}{u}^{3}-7\,{v}^{2}{w}^{3}u+270\,{u}^{6}-432\,{v}^{2}{u}^{4} -15\,{v}^{6}+162\,{v}^{4}{u}^{2}\geq 0$

This post has been edited 4 times. Last edited by xzlbq, Apr 21, 2017, 2:31 PM

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xzlbq

15849 posts

xzlbq

#14 h Apr 21, 2017, 10:30 PM • 2 Y

Y by vsathiam, Adventure10

Maverick wrote:

Prove that, for all positive real numbers $a$ , $b$ , $c$ , the inequality
$\frac {1}{a^3 + b^3 + abc} + \frac {1}{b^3 + c^3 + abc} + \frac {1}{c^3 + a^3 + abc} \leq \frac {1}{abc}$ holds.

Note

${a}^{3}+{b}^{3}+abc=ab \left( a+b+c \right) + \left( a+b \right) \left( a-b \right) ^{2}$
${\frac {1}{ab \left( a+b+c \right) }}+{\frac {1}{bc \left( a+b+c \right) }}+{\frac {1}{ca \left( a+b+c \right) }}={\frac {1}{abc}}$
Done.

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vsathiam

201 posts

vsathiam

#16 h Aug 22, 2017, 9:16 PM • 1 Y

Y by Adventure10

Not sure if this is what splintercell990 meant by bunching:

Clearing out the denominators and rearranging yields:

$\sum_{sym} a^6b^3 \geq 2\sum_{sym} a^5b^2c^2$

We can split this into two inequalities:

$\sum_{cyc} a^6b^3 \geq \sum_{sym} a^5b^2c^2$
and
$\sum_{cyc} a^6c^3 \geq \sum_{sym} a^5b^2c^2$

Both follow from weighted am-gm.

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sqing

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sqing

#33 h May 12, 2022, 9:57 AM

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Let $a, b, c$ be positive real numbers .Prove that
$\frac {a}{b + c +ka} + \frac {b}{c + a +kb}+\frac {c}{a + b +kc} \leq \frac{3}{k+2}$ Where $k\geq 1.$

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HamstPan38825

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HamstPan38825

#34 h Mar 15, 2023, 11:10 PM

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The key is to notice that $\frac{abc}{a^3+b^3+abc} \leq \frac c{a+b+c} \iff a^3+b^3 \geq a^2b + ab^2$ is true. Now just sum cyclically and divide by $abc$ .

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Quantum-Phantom

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Quantum-Phantom

#35 h Mar 16, 2023, 10:30 AM

Y by

Let $a^3=m$ , $b^3=n$ , $c^3=p$ , WLOG(homogeneous)let $mnp=1$ , we need to prove that
$\frac1{m+n+1}+\frac1{n+p+1}+\frac1{p+m+1}\le1,$ (expand) or $\sum(m^2n+mn^2)\ge2\sum m=2\sum m^{4/3}n^{1/3}p^{1/3}$ which is true by Murihead.

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megarnie

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megarnie

#36 h Jul 29, 2023, 4:10 PM • 1 Y

Y by OronSH

We see by Muirhead that $x^3 + y^3 \ge x^2y + xy^2$ for any positive reals $x,y$ . Hence $\begin{align*} \sum_{cyc} \frac{1}{a^3 + b^3 + abc} \\ \leq \sum_{cyc} \frac{1}{a^2b + ab^2 + abc} = \sum_{cyc} \frac{1}{ab(a + b + c)} \\ = \frac{1}{abc} ,\\ \end{align*}$ as desired.

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peppapig_

279 posts

peppapig_

#37 h Oct 1, 2023, 3:21 AM

Y by

WLOG, to homogenize, let $abc=1$ . Note that since by Muirhead's, we have that $a^2b+ab^2\geq a^3+b^3$ . Using this, we get that
$\sum_{\text{cyc}} \frac{1}{a^3+b^3+abc}=\sum_{\text{cyc}}\frac{abc}{a^3+b^3+abc} \leq \sum_{\text{cyc}} \frac{abc}{a^2b+ab^2+abc} = \sum_{\text{cyc}} \frac{c}{a+b+c}=1=\frac{1}{abc},$ finishing the problem.

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trk08

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trk08

#38 h Oct 15, 2023, 4:08 AM

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By Murihead's inequality:
$a^3+b^3\geq a^2b+ab^2.$ Therefore, the LHS is $\leq$ :
$\begin{align*} \sum_{\text{cyc}}\frac{1}{ab(a+b+c)}&=\frac{1}{a+b+c}\sum_{\text{cyc}}\frac{1}{ab}\\ &\leq\frac{a+b+c}{abc(a+b+c)}\\ &=\frac{1}{abc}.\\ \end{align*}$

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joshualiu315

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joshualiu315

#39 h Oct 24, 2023, 5:51 PM

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The equation is homogeneous so assume $abc=1$ . Then it suffices to prove

$\sum_{\text{cyc}} \frac{1}{a^3+b^3+abc} \le 1.$
Notice that $(3,0)$ majorizes $(2,1)$ so

$x^3+y^3 \ge x^2y+xy^2.$
by Muirhead. Thus, we have

$\begin{align*} \sum_{\text{cyc}} \frac{1}{a^3+b^3+abc} &= \sum_{\text{cyc}} \frac{abc}{a^3+b^3+abc} \\ &\le \sum_{\text{cyc}} \frac{abc}{a^2b+ab^2+abc} \\ &= \sum_{\text{cyc}} \frac{c}{a+b+c} = 1. \ \square \end{align*}$

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abc_978

6 posts

abc_978

#40 h Oct 26, 2023, 7:55 AM

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$a^3+b^3=(a+b)(a^2-ab+b^2)\geq(a+b)(ab)$ as $a^2-2ab+b^2\geq 0$ .

Thus $LHS\leq \sum_{\mathrm{cyc}}\frac{1}{ab(a+b+c)}= \frac{a+b+c}{abc(a+b+c)}= RHS$

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bjump

995 posts

bjump

#41 h Jan 24, 2024, 5:10 PM

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By AM-GM $a^{3}+b^{3} +abc \geq a^{2}b+ab^{2}+abc$ So $\frac{1}{ a^{2}b+ab^{2}+abc} \geq \frac{1}{a^{3}+b^{3}+abc}$
So
$\sum_{\text{cyc}} \frac{1}{a^{3}+b^{3}+abc} \leq \sum_{\text{cyc}} \frac{1}{a^{2}b+ab^{2}+abc}=\frac{1}{a+b+c}\sum_{\text{cyc}} \frac{1}{ab}= \frac{1}{abc}$ And we are done :gleam:

This post has been edited 1 time. Last edited by bjump, Jan 24, 2024, 5:10 PM

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Math4Life7

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Math4Life7

#42 h Feb 25, 2024, 10:43 PM

Y by

We multiply both sides by $(a^3+b^3+abc)(a^3+c^3+abc)(b^3+c^3+abc)abc$ . We cancel terms and get $\sum_{\text{sym}} a^6b^3 \geq \sum_{\text{sym}} a^5b^2c^2$ This is obvious from Muirhead. $\blacksquare$

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ihatemath123

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ihatemath123

#43 h May 3, 2024, 3:30 AM

Y by

WLOG set $abc = 1$ , so the inequality becomes
$\frac{1}{a^3 + b^3 + 1} + \frac{1}{b^3+c^3 + 1} + \frac{1}{c^3 + a^3 + 1} \leq 1.$ Expanding both sides and manipulating, the inequality simplifies (reversibly) to
$\begin{align*} 2(a^3 + b^3 +c^3) & \leq \sum_{\text{sym}} a^6b^3 \\ 2 \sum_{\text{cyc}}a^5 b^2 c^2 & \leq \sum_{\text{sym}} a^6b^3, \end{align*}$ which is true by Muirhead's theorem.

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blueprimes

314 posts

blueprimes

#44 h Jun 12, 2024, 11:10 AM

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Without loss of generality, assume $abc = 1$ , then we wish to show
$(a^3 + b^3 + 1)(b^3 + c^3 + 1) + (b^3 + c^3 + 1)(c^3 + a^3 + 1) + (c^3 + a^3 + 1)(a^3 + b^3 + 1) = (a^3 + b^3 + 1)(b^3 + c^3 + 1)(c^3 + a^3 + 1).$ $\iff \sum a^6 + 3 \sum a^3b^3 + 4 \sum a^3 + 3 \le 2a^3b^3c^3 + \sum a^6b^3 + \sum a^6 + 3 \sum a^3b^3 + 2 \sum a^3 + 1$ $\iff 2 \sum a^3 \le \sum a^6b^3$ $\iff 2 \sum a^5 b^2 c^2 \le \sum a^6 b^3$ which follows by summing the Weighted AM-GM
$\frac{2}{3} a^6b^3 + \frac{1}{3} c^6a^3 \ge \sqrt[3]{a^{15}b^6c^6} = a^5b^2c^2$ symmetrically.

This post has been edited 2 times. Last edited by blueprimes, Jun 12, 2024, 11:12 AM

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The_Eureka

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The_Eureka

#45 h Oct 3, 2024, 9:10 AM

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By AM-GM, $a ^ 3 + b ^ 3 + abc \geq 3a ^ {4 / 3}b ^ {4/3}c ^ {1/3}$ .
Thus, $\text{LHS} \leq \sum_{\text{cyc}} \frac{1}{3} a ^ {-4/3}b ^ {-4/3}c ^ {-1/3}.$
It now suffices to prove that
$\begin{align*} \frac{1}{3}\sum_{\text{cyc}}a ^ {-4/3}b ^ {-4/3}c ^ {-1/3} &\leq \frac{1}{abc} \\ \iff \sum_{\text{cyc}}a ^ {-4/3}b ^ {-4/3}c ^ {-1/3} &\leq \frac{3}{abc} \\ \iff \sum_{\text{sym}}a ^ {-4/3}b ^ {-4/3}c ^ {-1/3} &\leq \sum_{\text{sym}}a ^ {-1}b ^ {-1}c ^ {-1}, \end{align*}$ which follows from $(-1, -1, -1) \succ (-4/3, -4/3, -1/3)$ and Muirhead's Inequality.

This post has been edited 1 time. Last edited by The_Eureka, Oct 3, 2024, 9:10 AM

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megahertz13

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megahertz13

#46 h Nov 3, 2024, 1:44 AM • 1 Y

Y by kilobyte144

The main idea is that $a^3+b^3\ge a^2b+b^2a.$ Note that $\frac{1}{a^3+b^3+abc} + \frac{1}{b^3+c^3+abc} + \frac{1}{c^3+a^3+abc}$ $\le \frac{1}{a^2b+b^2a+abc}+\frac{1}{b^2c+c^2b+abc}+\frac{1}{c^2a+a^2c+abc}$ $= (\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca})\cdot \frac{1}{a+b+c}$ $= \frac{a+b+c}{abc}\cdot \frac{1}{a+b+c} = \frac{1}{abc}.$