Prove that, for all positive real numbers a, b, c, the inequality 1a3+b3+abc+1b3+c3+abc+1c3+a3+abc≤1abc 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
a3+b3≥ab2+a2b, so we get that: ∑abca3+b3+abc≤∑abcab(a+b+c)=∑ca+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 aabc(a+b+c), etc. Solution. And miraculously it works. Of course by muirhead and [3,0]≻[2,1] one has a3+b3≥a2b+ab2. Then, the left hand side is at most ∑cycaabc(a+b+c)=1abc.
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 a2bc+b2ca≥(a+b)2c(a+b)⇒a2bc+b2ca+1≥a+b+cc ⇒1a2bc+b2ca+1≤ca+b+c⇒∑cyc1a2bc+b2ca+1≤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 1a3+b3+abc+1b3+c3+abc+1c3+a3+abc≤1abcholds. Prove that, for all positive real numbers a, b, c, the inequality 1a3+b3+abc+1b3+c3+abc+1c3+a3+abc≥52abc+a3+b3+c3holds.
21.04.2017 17:27
xzlbq wrote: Prove that, for all positive real numbers a, b, c, the inequality 1a3+b3+abc+1b3+c3+abc+1c3+a3+abc≥52abc+a3+b3+c3holds. <=> 27u6−54v2u4+10w3u3+27v4u2−11v2w3u+v6+w627w3u6−81v2w3u4+27v6u3+11w6u3+54v4w3u2−15v2w6u−27v8u+w9+4v6w3≥527u3−27uv2+5w3 <=> 27w3(22w3u3−7v2w3u+270u6−432v2u4−15v6+162v4u2)+729u(u−v)(u+v)(3u2−4v2)(−v2+3u2)2≥0 Need to prove 22w3u3−7v2w3u+270u6−432v2u4−15v6+162v4u2≥0
22.04.2017 01:30
Maverick wrote: Prove that, for all positive real numbers a, b, c, the inequality 1a3+b3+abc+1b3+c3+abc+1c3+a3+abc≤1abcholds. Note a3+b3+abc=ab(a+b+c)+(a+b)(a−b)2 1ab(a+b+c)+1bc(a+b+c)+1ca(a+b+c)=1abc Done.
23.08.2017 00:16
Not sure if this is what splintercell990 meant by bunching: Clearing out the denominators and rearranging yields: ∑syma6b3≥2∑syma5b2c2 We can split this into two inequalities: ∑cyca6b3≥∑syma5b2c2 and ∑cyca6c3≥∑syma5b2c2 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 1a3+b3+abc+1b3+c3+abc+1c3+a3+abc≤1abcholds. stronger (x2+y2+z2)2(x+y+z)(x3+y3+z3)≥xyzx3+y3+xyz+xyzy3+z3+xyz+xyzz3+x3+xyz (x3+y3+z3)2(x2+y2+z2)(x4+y4+z4)≥xyzx3+y3+xyz+xyzy3+z3+xyz+xyzz3+x3+xyz (x+y+z)(x2+y2+z2)xyz(xy2+yz2+x2z)(x2y+y2z+z2x)≥xyzx3+y3+xyz+xyzy3+z3+xyz+xyzz3+x3+xyz
23.08.2017 02:46
have this: 12≥y3z3y6+z6+4x2y2z2+x3z3z6+x6+4x2y2z2+x3y3x6+y6+4x2y2z2
23.08.2017 02:51
a,b,c be sides of a triangle,prove b3c3b6+c6+29a2b2c2+c3a3c6+a6+29a2b2c2+a3b3a6+b6+29a2b2c2≥331
18.09.2017 09:26
Clearing the denominator, we get S(5,2,2)≤S(6,3,0) This is just Muirhead inequality
30.01.2018 06:50
By CS we have, (a3+b3+abc)(1a+1b+cab)≥(a+b+c)2⟹1a3+b3+abc≤cabc(a+b+c)Summing the three inequalities, we get ∑cyc1a3+b3+abc≤∑cyccabc(a+b+c)=1abc
30.01.2018 07:24
lets use the following inequality a3+b3≥ab(a+b) then ∑cyc1a3+b3+abc≤∑cyc1ab(a+b)+abc≤1abc
30.01.2018 08:56
Arne wrote: We havex3+y3≥x2y+xy2 for all x,y>0 by rearrangement. So ∑1(a3+b3+abc)≤∑1(a2b+ab2+abc)=1abc 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^2is 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^2This 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^3which 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^2symmetrically.
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}.