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.
Problem
Source: USAMO 1997/5; also: ineq E2.37 in Book: Inegalitati; Authors:L.Panaitopol,V. Bandila,M.Lascu
Tags: inequalities, AMC, USA(J)MO, USAMO, symmetry, three variable inequality, Hi
12.09.2003 19:04
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.
14.03.2005 01:38
$a^3+b^3\geq ab^2+a^2b$, so we get that: $\sum \frac{abc}{a^3+b^3+abc}\leq \sum\frac{abc}{ab(a+b+c)} = \sum\frac{c}{a+b+c}=1$
14.03.2005 03:24
Ok, thanks.
14.03.2005 03:51
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.
14.03.2005 04:03
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.
25.07.2005 06:27
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 ]
15.01.2006 05:34
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:
18.03.2016 05:04
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 $\sum_{cyc} \frac{a}{abc(a+b+c)} = \frac{1}{abc}.$
16.05.2016 09:08
Does this count as a one liner ?
21.04.2017 14:22
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 $a=b=c$.
21.04.2017 16:26
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.
21.04.2017 17:27
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 \[22\,{w}^{3}{u}^{3}-7\,{v}^{2}{w}^{3}u+270\,{u}^{6}-432\,{v}^{2}{u}^{4} -15\,{v}^{6}+162\,{v}^{4}{u}^{2}\geq 0\]
22.04.2017 01:30
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.
23.08.2017 00:16
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.
23.08.2017 02:25
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. stronger \[{\frac { \left( {x}^{2}+{y}^{2}+{z}^{2} \right) ^{2}}{ \left( x+y+z \right) \left( {x}^{3}+{y}^{3}+{z}^{3} \right) }}\geq {\frac {xyz}{{x}^{ 3}+{y}^{3}+xyz}}+{\frac {xyz}{{y}^{3}+{z}^{3}+xyz}}+{\frac {xyz}{{z}^{ 3}+{x}^{3}+xyz}}\] \[{\frac { \left( {x}^{3}+{y}^{3}+{z}^{3} \right) ^{2}}{ \left( {x}^{2}+ {y}^{2}+{z}^{2} \right) \left( {x}^{4}+{y}^{4}+{z}^{4} \right) }}\geq { \frac {xyz}{{x}^{3}+{y}^{3}+xyz}}+{\frac {xyz}{{y}^{3}+{z}^{3}+xyz}}+{ \frac {xyz}{{z}^{3}+{x}^{3}+xyz}}\] \[{\frac { \left( x+y+z \right) \left( {x}^{2}+{y}^{2}+{z}^{2} \right) xyz}{ \left( x{y}^{2}+y{z}^{2}+{x}^{2}z \right) \left( {x}^{2}y+{y}^{ 2}z+{z}^{2}x \right) }}\geq {\frac {xyz}{{x}^{3}+{y}^{3}+xyz}}+{\frac {xyz }{{y}^{3}+{z}^{3}+xyz}}+{\frac {xyz}{{z}^{3}+{x}^{3}+xyz}}\]
23.08.2017 02:46
have this: \[\frac{1}{2}\geq {\frac {{y}^{3}{z}^{3}}{{y}^{6}+{z}^{6}+4\,{x}^{2}{y}^{2}{z}^{2}}} +{\frac {{x}^{3}{z}^{3}}{{z}^{6}+{x}^{6}+4\,{x}^{2}{y}^{2}{z}^{2}}}+{ \frac {{x}^{3}{y}^{3}}{{x}^{6}+{y}^{6}+4\,{x}^{2}{y}^{2}{z}^{2}}}\]
23.08.2017 02:51
$a,b,c$ be sides of a triangle,prove \[{\frac {{b}^{3}{c}^{3}}{{b}^{6}+{c}^{6}+29\,{a}^{2}{b}^{2}{c}^{2}}}+{ \frac {{c}^{3}{a}^{3}}{{c}^{6}+{a}^{6}+29\,{a}^{2}{b}^{2}{c}^{2}}}+{ \frac {{a}^{3}{b}^{3}}{{a}^{6}+{b}^{6}+29\,{a}^{2}{b}^{2}{c}^{2}}}\geq { \frac {3}{31}}\]
18.09.2017 09:26
Clearing the denominator, we get $S(5,2,2) \leq S(6,3,0)$ This is just Muirhead inequality
30.01.2018 06:50
By CS we have, \[(a^3+b^3+abc)\left(\frac1a+\frac1b+\frac{c}{ab}\right)\geq (a+b+c)^2\]\[\implies \frac{1}{a^3+b^3+abc}\leq \frac{c}{abc(a+b+c)}\]Summing the three inequalities, we get \[\sum_{cyc} \frac{1}{a^3+b^3+abc} \leq \sum_{cyc} \frac{c}{abc(a+b+c)} = \frac{1}{abc}\]
30.01.2018 07:24
lets use the following inequality $a^3+b^3 \geq ab(a+b)$ then $\sum_{cyc} \frac{1}{a^3+b^3+abc} \leq \sum_{cyc} \frac{1}{ab(a+b)+abc} \leq \frac {1}{abc}$
30.01.2018 08:56
Arne wrote: We have$ x^3 + y^3 \geq x^2y + xy^2$ for all $ x, y > 0$ by $ rearrangement.$ So $\sum \dfrac {1}{(a^3 + b^3 + abc)} \le \sum \dfrac {1}{(a^2b + ab^2 + abc)} = \dfrac {1}{abc} $ by simple computation. Fixed Arne's solution
30.01.2018 09:55
$a^2–2ab+b^2\ge 0$ $\implies a^2–ab+b^2\ge ab$ $\implies (a+b)(a^2–ab+b^2)\ge (a+b)ab$ $a^3+b^3\ge ab(a+b)$ $\implies ab(a+b)+abc=ab(a+b+c)$ And thus follows I think everyone has the same conclusion.............
07.05.2020 03:53
We know that $a, b, c > 0$. By Chebyshev's inequality we have $(\sum_{cyc}\frac{1}{a^3+b^3+abc})(2(a^3+b^3+c^3) + 3abc) \leq 9$ Notice that $(2(a^3+b^3+c^3) + 3abc) \leq 3(a^3+b^3+c^3)$ by AM GM $(\sum_{cyc}\frac{1}{a^3+b^3+abc}) \leq \frac{3}{a^3+b^3+c^3}$ By AM GM $a^3+b^3+c^3 \geq 3abc$ so it follows that $(\sum_{cyc}\frac{1}{a^3+b^3+abc}) \leq \frac{3}{a^3+b^3+c^3} \leq \frac{1}{abc}$. $\blacksquare$
07.05.2020 04:00
Wait in the Chebyshev's Inequality part, shouldn't it be the other way around, because Cauchy Scwarz says that is greater than or equal to 9???
30.01.2021 07:06
Solved with nprime06 Since the inequality is homogenous, WLOG let $a+b+c = 1$ and since $a^3+b^3 \ge a^2b+b^2a$ we have \[\sum_{\text{cyc}} \frac{1}{a^3+b^3+abc} \le \sum_{\text{cyc}} \frac{1}{a^2b+b^2a+abc} = \frac{1}{a+b+c}\sum_{\text{cyc}} \frac{1}{ab} = \sum_{\text{cyc}} \frac{1}{ab} \le \frac{1}{abc} \]which is trivial by expanding. $\square$
30.01.2021 07:32
fukano_2 wrote: Solved with nprime06 Since the inequality is homogenous, WLOG let $a+b+c = 1$ and since $a^3+b^3 \ge a^2b+b^2a$ we have \[\sum_{\text{cyc}} \frac{1}{a^3+b^3+abc} \le \sum_{\text{cyc}} \frac{1}{a^2b+b^2a+abc} = \frac{1}{a+b+c}\sum_{\text{cyc}} \frac{1}{ab} = \sum_{\text{cyc}} \frac{1}{ab} \le \frac{1}{abc} \]which is trivial by expanding. $\square$ Why do you even need to hold that ? pls correct me if I am wrong . Quote: let $a+b+c=1$ Since by Muirhead Inequality, \[\sum_{\text{cyc}} \frac{1}{a^3+b^3+abc} \le \sum_{\text{cyc}} \frac{1}{a^2b+b^2a+abc} \]$$\sum_{\text{cyc}} \frac{1}{a^2b+b^2a+abc}=\frac{1}{a^2b+b^2a+abc}+\frac{1}{b^2c+c^2b+abc}+\frac{1}{c^2a+a^2c+abc}$$$$=\frac{1}{a+b+c}\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)$$$$\left(\frac{1}{a+b+c}\right)\left(\frac{a+b+c}{abc}\right)=\frac{1}{abc}$$So it true that, $$ \frac {1}{a^3 + b^3 + abc} + \frac {1}{b^3 + c^3 + abc} + \frac {1}{c^3 + a^3 + abc} \leq \frac {1}{abc} $$Proved $\square$
03.07.2021 18:27
Here is mine solution just for storage and make things simple if something wrong in it please tell me. My solution: Here we can use that \[a^3+b^3\geq ab(a+b)\]this can be seen by the $(a+b)^3$ formula and proved by muirhead because we know that $(3,0)\succ(2,1)$ Now, We can see that \[\sum_{cyc}\frac{1}{a^3+b^3+abc}\leq \sum_{cyc}\frac{1}{ab(a+b)+abc}=\sum_{cyc}\frac{1}{ab(a+b+c)}\]After multiplying but $c$ in numerator and denominator we can get\[\sum_{cyc}\frac{1}{a^3+b^3+abc}\leq \sum_{cyc}\frac{c}{abc(a+b+c}=\frac{1}{abc}\sum_{cyc}\frac{c}{a+b+c}=\frac{1}{abc}\]And we are done !
12.07.2021 21:21
By Rearrangement, $a^3+b^3\ge a^2b+ab^2$. Then we homogenize by assuming $a+b+c=1$. Finally: $$\sum_{\text{cyc}}\frac1{a^3+b^3+abc}\le\sum_{\text{cyc}}\frac1{a^2b+ab^2+abc}=\sum_{\text{cyc}}\frac1{ab}=\frac1{abc}.~\square$$
12.05.2022 07:16
Let $ a, b, c$ be positive real numbers .Prove that $$ \frac{1}{a^3+b^3+kabc} + \frac{1}{b^3+c^3+kabc} + \frac{1}{a^3+c^3+kabc} \leq \frac{3}{(k+2)abc}$$Where $k\geq 1.$
12.05.2022 07:39
blahblahblah wrote: $a^3+b^3\geq ab^2+a^2b$, so we get that: $\sum \frac{abc}{a^3+b^3+abc}\leq \sum\frac{abc}{ab(a+b+c)} = \sum\frac{c}{a+b+c}=1$ very effective
12.05.2022 12:57
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 $k\geq 1.$
16.03.2023 02:10
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 $abc$.
16.03.2023 13:30
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.
29.07.2023 19:10
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.
01.10.2023 06:21
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.
15.10.2023 07:08
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*}
24.10.2023 20:51
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*}
26.10.2023 10:55
$a^3+b^3=(a+b)(a^2-ab+b^2)\geq(a+b)(ab)$ as $a^2-2ab+b^2\geq 0$. Thus $LHS\leq \sum_{\mathrm{cyc}}\frac{1}{ab(a+b+c)}= \frac{a+b+c}{abc(a+b+c)}= RHS$
24.01.2024 20:10
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
26.02.2024 01:43
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. $\blacksquare$
03.05.2024 06:30
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.
12.06.2024 14:10
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.
03.10.2024 12:10
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.
03.11.2024 04:44
The main idea is that $a^3+b^3\ge a^2b+b^2a.$ Note that $$\frac{1}{a^3+b^3+abc} + \frac{1}{b^3+c^3+abc} + \frac{1}{c^3+a^3+abc}$$$$\le \frac{1}{a^2b+b^2a+abc}+\frac{1}{b^2c+c^2b+abc}+\frac{1}{c^2a+a^2c+abc}$$$$ = (\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca})\cdot \frac{1}{a+b+c}$$$$ = \frac{a+b+c}{abc}\cdot \frac{1}{a+b+c} = \frac{1}{abc}.$$