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

• KLscilevothma
In summary: So, in summary, the given equation 1/[a3(b+c)] + 1/[b3(a+c)] + 1/[c3(b+a)] >= 3/2 can be proven using the AM>GM inequality, but there is a mistake in the solution provided. In order to obtain the desired result, the substitution must be made in such a way that the expression becomes greater than 3/2, not less than 3/2. Additionally, the given equation is only true for the case where a, b, and c are all equal to 1.
KLscilevothma
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.

Last edited by a moderator:
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...

Last edited:
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"...

Last edited by a moderator:
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!

Last edited:
No...there are only a few "feasable" substitutions...x=1/a...x=a+b...x=a-b...things like these...

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 ?

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...

I think there must be something wrong in the proof, we can't get logically correct answers which contratict each other.

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 ?

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.

Last edited:
No...
We took E>X, and then expected to prove X>3/2...well..this is not true for every X...got it ?

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

[/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

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.

Last edited:
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

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.

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/"

Last edited by a moderator:

## 1. What is the inequality being proven?

The inequality being proven is 1/[a3(b+c)] + 1/[b3(a+c)] + 1/[c3(b+a)] ≥ 3/2.

## 2. What is the hint for proving this inequality?

The hint is to use the AM-GM (Arithmetic Mean-Geometric Mean) inequality.

## 3. How does the AM-GM inequality help in proving this inequality?

The AM-GM inequality states that for positive real numbers, the arithmetic mean is always greater than or equal to the geometric mean. This can be applied to the terms in the given inequality to simplify it and prove that it is greater than or equal to 3/2.

## 4. What are the necessary conditions for the AM-GM inequality to hold?

The necessary conditions for the AM-GM inequality to hold are that all terms must be positive and there must be an even number of terms.

## 5. Can this inequality be proven using other methods besides the hint provided?

Yes, there are other mathematical techniques that can be used to prove this inequality, such as Cauchy-Schwarz inequality, rearrangement inequality, or the method of Lagrange multipliers.

• General Math
Replies
3
Views
1K
• General Math
Replies
1
Views
881
• General Math
Replies
1
Views
837
• General Math
Replies
1
Views
887
• General Math
Replies
2
Views
1K
• General Math
Replies
2
Views
791
• General Math
Replies
1
Views
1K
• General Math
Replies
8
Views
1K
• General Math
Replies
1
Views
611
• General Math
Replies
1
Views
640