Prove Inequality 1/[a3(b+c)] + 1/[b3(a+c)] + 1/[c3(b+a)] >= 3/2 with Hint

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Discussion Overview

The discussion revolves around proving the inequality 1/[a³(b+c)] + 1/[b³(a+c)] + 1/[c³(a+b)] ≥ 3/2 under the condition that abc=1, where a, b, and c are positive real numbers. Participants explore various mathematical inequalities and approaches to tackle the problem, including the AM-GM inequality and the Cauchy-Buniakowsky-Schwartz inequality.

Discussion Character

  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant suggests using the AM-GM inequality to establish a lower bound for the sum of the fractions.
  • Another participant argues that the product of the terms leads to a contradiction, as they derive (a+b)(a+c)(b+c) ≥ 8 instead of the required ≤ 8.
  • Some participants express confusion over the application of inequalities, questioning how both (a+b)(a+c)(b+c) can be shown to be both ≥ 8 and ≤ 8.
  • A later reply proposes that the only solution occurs when a = b = c = 1, leading to the conclusion that the inequality holds true in this specific case.
  • Several participants share their experiences with similar problems in math contests, indicating a broader context of competitive mathematics.
  • One participant mentions the Cauchy-Buniakowsky-Schwartz inequality and its relation to the problem, seeking clarification on its application.

Areas of Agreement / Disagreement

Participants generally disagree on the application of inequalities and the validity of certain steps in the proof. There is no consensus on the correct approach or solution to the problem, as multiple viewpoints and interpretations are presented.

Contextual Notes

Participants note the importance of the positivity condition for a, b, and c, which was initially overlooked by some. The discussion also highlights the complexity of applying inequalities correctly, with some steps leading to contradictory results.

Who May Find This Useful

Readers interested in advanced mathematical inequalities, competitive mathematics, or those preparing for math contests may find this discussion relevant.

KLscilevothma
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If abc=1
prove
1/[a3(b+c)] + 1/[b3(a+c)] + 1/[c3(b+a)] >= 3/2
where a, b, c are positive real numbers.

Please give me a hint.
 
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I think that a>0, b>0, c>0...because if this is not true then for a=5, b=-1, c=-1/5 the left part is equal to -31.465, which is smaller than 3/2...
So...I'll use this well known inequality (I don't know the english name)...
(a1+a2+..+an)/n>=(a1*a2*...*an)^(1/n)...where a1,...an>0 and n>=2...
Let n=3 and a1=1/(a^3*(b+c)), a2=1/(b^3*(a+c)), a3=1/(c^3*(a+b))...
We obtain (a1+a2+a3)/3>(a1*a2*a3)^(1/3)...

Let S=a1+a2+a3...
S/3>(a1*a2*a3)^(1/3)...evidently a1*a2*a3=1/[(a+b)*(a+c)*(b+c)]...because a^3*b^3*c^3=1...
So...S/3>(1/[(a+b)*(a+c)*(b+c)])^(1/3)...
S>3*(1/[(a+b)*(a+c)*(b+c)])^(1/3)...
It's simple to prove now that (1/[(a+b)*(a+c)*(b+c)])^(1/3)>1/2...the same inequality I used before...good luck...
 
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wow bogdan, thank you!

yes, a, b, c are positive real numbers and I forgot to include that in the question.

So...I'll use this well known inequality (I don't know the english name)...
(a1+a2+..+an)/n>=(a1*a2*...*an)^(1/n)...where a1,...an>0 and n>=2...

I only call it AM>=GM.

Let n=3 and a1=1/(a^3*(b+c)), a2=1/(b^3*(a+c)), a3=1/(c^3*(a+b))...
We obtain (a1+a2+a3)/3>(a1*a2*a3)^(1/3)...

Let S=a1+a2+a3...
S/3>(a1*a2*a3)^(1/3)...evidently a1*a2*a3=1/[(a+b)*(a+c)*(b+c)]...because a^3*b^3*c^3=1...
So...S/3>(1/[(a+b)*(a+c)*(b+c)])^(1/3)...
S>3*(1/[(a+b)*(a+c)*(b+c)])^(1/3)...
It's simple to prove now that (1/[(a+b)*(a+c)*(b+c)])^(1/3)>=1/2

prove 1/[(a+b)*(a+c)*(b+c)])^(1/3)>=1/2
==> Prove (a+b)*(a+c)*(b+c) <=8
use AM>=GM again
(a+b)/2 >=sqrt(ab)...(1)
(a+c)/2 >=sqrt(ac)...(2)
(c+b)/2 >=sqrt(cb)...(3)
(1)*(2)*(3)
the result follows
 
Unfortunately...it is so very wrong...
(a+b)*(a+c)*(b+c)>=8, not <=8...
There is a solution...use Cauchy-Buniakowsky-Schwartz inequaltity and then AM>GM...
 
This problem is very strange...I strongly believe I've seen it somewhere in a contest...in the Balcaniad math contest...or something...but that sugestion (Cauchy inequality and AM>GM) is good...because I have it in a book...and that's the only hint the author gives...
 
If I remember correctly, it's an IOM question, 1995.

I think the Cauchy-Buniakowsky-Schwartz inequaltity is different from Cauchy-Schwartz inequality. I haven't heard of C-B-S inequality before. Do you have any link or some explanations about what CBS inequality is?
 
No...the two are the same...
IOM ? Are you crazy ? Do you know what a headache this inequality gave me ?
I can't do much more simple inequalities...and I tried this...
I have this problem in a book for preparing math contests...and the explanation is : use cauchy-...-...- inequality and AM>GM...that's all...and...what a luck...I have a book with IOMs but 1990-1994...
 
well, frankly speaking, I needed to take out my textbook before I could remember what C-S inequality is!

hehe, I'm not that good in mathematics. I just saw this question and thought it might be an interesting one. I think I got the solutions of IMO questions in 1995 somewhere in my computer but I was too lazy to find it.

I can't do much more simple inequalities
same here.

I need to stop here now, and revise for my pure math test today!
Perhaps we can do a little discussion on inequalities later.
 
Here's the solution...problem number 2...

http://imo.math.ca/Sol/95/95_prob.html



It's so "simple"...
 
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  • #10
usually applying substitutions in harder inequality problems can simplify the questions a bit, but to find suitable substitions is difficult.

yes, the solution is so nice!
 
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  • #11
No...there are only a few "feasable" substitutions...x=1/a...x=a+b...x=a-b...things like these...
 
  • #12
prove 1/[(a+b)*(a+c)*(b+c)])^(1/3)>=1/2
==> Prove (a+b)*(a+c)*(b+c) <=8
use AM>=GM again
(a+b)/2 >=sqrt(ab)...(1)
(a+c)/2 >=sqrt(ac)...(2)
(c+b)/2 >=sqrt(cb)...(3)
(1)*(2)*(3)

what's wrong in the above steps ?
 
  • #13
We have to prove (a+b)*(a+c)*(b+c)<=8, but using AM>GM we obtain (a+b)*(a+c)*(b+c)>=8...exactly the opposite...I've seen another inequality like this...in a IMO romanian team selection test...where after applying AM>GM twice I obtained the opposite inequality...and took 0 points...
 
  • #14
I think there must be something wrong in the proof, we can't get logically correct answers which contratict each other.
 
  • #15
I've just shown you the mistake...(a+b)*(b+c)*(a+c)>=8, not viceversa...we wanted the opposite inequality...there's the mistake...we made the expression too small...and now the new expression is less than 3/2...got it ?
 
  • #16
We have to prove (a+b)*(a+c)*(b+c)<=8, but using AM>GM we obtain (a+b)*(a+c)*(b+c)>=8...exactly the opposite

Bogdan, I think you've misunderstood me or I wasn't explain my question clearly. I meant how come we can prove (a+b)*(b+c)*(a+c)>=8 and (a+b)*(b+c)*(a+c)<=8 at the same time ? (I know we have to prove (a+b)*(a+c)*(b+c)<=8, not the other way round.) Like we get the wrong answer if we use AM>=GM while it is ok when using other methods. So what's wrong if we use AM>=GM?


If the question is modified a bit, say if abc=1 (a,b,c are positive real numbers)
which of the following is correct
1/[a3(b+c)] + 1/[b3(a+c)] + 1/[c3(b+a)] >= 3/2
or
1/[a3(b+c)] + 1/[b3(a+c)] + 1/[c3(b+a)] <= 3/2

we can't have z>=3/2 and z<=3/2 at the same time while z is not necessarily equals 3/2.
 
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  • #17
No...
We took E>X, and then expected to prove X>3/2...well..this is not true for every X...got it ?
 
  • #18


Originally posted by KL Kam
If abc=1
prove
1/[a3(b+c)] + 1/[b3(a+c)] + 1/[c3(b+a)] >= 3/2

Please give me a hint. [/QUOTE

a=1 b=1 c=1 this is the only possible solution if the latter equation is tru. the only other way for three variables to multiply to equal 1 is to have two of them equal -1 but if that is true then u get 1/0+1/0+1/0 and this is not possible. there for if you replace every variable with 1 you get 1/2+1/2+1/2 which equals 3/2
 
  • #19
something must have gone wrong in our proof.

It's simple to prove now that (1/[(a+b)*(a+c)*(b+c)])^(1/3)>=1/2
This step is wrong, which leads the rest of our proof a contradiction. (one of the counter examples, a=1.1, b=2.2, c=3.3)
Originally Posted by true_omega
the only other way for three variables to multiply to equal 1 is to have two of them equal -1
I'm sorry,true_omega, that I didn't include that a, b, c are positive real numbers at the very beginning. I've edited it already.
 
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  • #20
i did not say the only way is for two of the variables to equal -1 i said the only other way. the solution would be for each variable to equal positive 1 whereas 1*1*1 of course equals 1
 
  • #21
As a brand spankin' new member here, I look at this post and think to myself, THIS is how to use the internet.

Very happy to be here, I was here a LONG time ago, but that was before I had taken physics, and before I was much a science-person so now I will have a lot more in common.
 
  • #22
An alternative solution

Please visit the following page for an alternative proof.
http://texasmath.org/forum/viewtopic.php?f=16&t=432&start=0&st=0&sk=t&sd=a"










http://www.idealmath.com/"
 
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