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Inequality If abc=1

by KLscilevothma
Tags: abc1, inequality
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KLscilevothma
#1
Apr30-03, 07:25 PM
P: 321
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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bogdan
#2
May1-03, 04:26 AM
P: 192
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...
KLscilevothma
#3
May1-03, 05:02 AM
P: 321
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

bogdan
#4
May1-03, 07:09 AM
P: 192
Inequality If abc=1

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...
bogdan
#5
May1-03, 09:46 AM
P: 192
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...
KLscilevothma
#6
May1-03, 03:39 PM
P: 321
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 explainations about what CBS inequality is?
bogdan
#7
May1-03, 03:43 PM
P: 192
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...
KLscilevothma
#8
May1-03, 03:52 PM
P: 321
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.
bogdan
#9
May1-03, 03:54 PM
P: 192
Here's the solution...problem number 2...

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



It's so "simple"...
KLscilevothma
#10
May1-03, 05:11 PM
P: 321
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!
bogdan
#11
May2-03, 05:19 AM
P: 192
No...there are only a few "feasable" substitutions...x=1/a...x=a+b...x=a-b...things like these...
KLscilevothma
#12
May2-03, 07:41 PM
P: 321
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 ?
bogdan
#13
May3-03, 03:49 AM
P: 192
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...
KLscilevothma
#14
May3-03, 04:29 AM
P: 321
I think there must be something wrong in the proof, we can't get logically correct answers which contratict each other.
bogdan
#15
May3-03, 11:28 AM
P: 192
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 ?
KLscilevothma
#16
May4-03, 12:41 AM
P: 321
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.
bogdan
#17
May5-03, 09:34 AM
P: 192
No...
We took E>X, and then expected to prove X>3/2......well..this is not true for every X...got it ?
true_omega
#18
May5-03, 08:37 PM
P: n/a
[QUOTE]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


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