# Inequality If abc=1

by KLscilevothma
Tags: abc1, inequality
 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.
 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...[6)]
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. [:D]

 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

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...
 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...
 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?
 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... [s(]
P: 321
well, frankly speaking, I needed to take out my textbook before I could remember what C-S inequality is! [g)]

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.
 P: 192 Here's the solution...problem number 2... http://imo.math.ca/Sol/95/95_prob.html [:D] It's so "simple"...[s(]
 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!
 P: 192 No...there are only a few "feasable" substitutions...x=1/a...x=a+b...x=a-b...things like these...[t)]
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 ?
 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...[a)]
 P: 321 I think there must be something wrong in the proof, we can't get logically correct answers which contratict each other.
 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 ?[8)]
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.
 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 ?
 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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