
#1
Apr3003, 07:25 PM

P: 321

If abc=1
prove 1/[a^{3}(b+c)] + 1/[b^{3}(a+c)] + 1/[c^{3}(b+a)] >= 3/2 where a, b, c are positive real numbers. Please give me a hint. 



#2
May103, 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...[6)] 



#3
May103, 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. ==> 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 



#4
May103, 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 CauchyBuniakowskySchwartz inequaltity and then AM>GM... 



#5
May103, 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...




#6
May103, 03:39 PM

P: 321

If I remember correctly, it's an IOM question, 1995.
I think the CauchyBuniakowskySchwartz inequaltity is different from CauchySchwartz inequality. I haven't heard of CBS inequality before. Do you have any link or some explainations about what CBS inequality is? 



#7
May103, 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 19901994... [s(] 



#8
May103, 03:52 PM

P: 321

well, frankly speaking, I needed to take out my textbook before I could remember what CS 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 need to stop here now, and revise for my pure math test today! Perhaps we can do a little discussion on inequalities later. 



#9
May103, 03:54 PM

P: 192

Here's the solution...problem number 2...
http://imo.math.ca/Sol/95/95_prob.html [:D] It's so "simple"...[s(] 



#10
May103, 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! 



#11
May203, 05:19 AM

P: 192

No...there are only a few "feasable" substitutions...x=1/a...x=a+b...x=ab...things like these...[t)]




#12
May203, 07:41 PM

P: 321





#13
May303, 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...[a)]




#14
May303, 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.




#15
May303, 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 ?[8)]




#16
May403, 12:41 AM

P: 321

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/[a^{3}(b+c)] + 1/[b^{3}(a+c)] + 1/[c^{3}(b+a)] >= 3/2 or 1/[a^{3}(b+c)] + 1/[b^{3}(a+c)] + 1/[c^{3}(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. 



#17
May503, 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 ? 


#18
May503, 08:37 PM

P: n/a

[QUOTE]Originally posted by KL Kam
If abc=1 prove 1/[a^{3}(b+c)] + 1/[b^{3}(a+c)] + 1/[c^{3}(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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