Let $a,b,c$ be positive reals s.t. $a^2+b^2+c^2=1$. Prove the following inequality \[ \sum \frac{a}{a^3+bc} >3 . \] Proposed by A. Khrabrov
Problem
Source: Tuymaada, Day 2, Problem 8
Tags: inequalities, inequalities proposed
31.07.2005 05:50
I know one member in this forum can answer it very well. I saw him sovle this problem . His solution is really great, natural and really nice. He is Ng.T.TU. Hey TU write your solution right now!!!!!! I want to post that solution but It is not mine. So I don't want the others think I stole the solution from him. I am a new member. Hi everyone!!!!!!!!!!
01.08.2005 06:24
N.T.TU will never post his proof unless you say you hate it.
01.08.2005 11:44
Yes I really hate it!!!
01.08.2005 12:14
Somehow I don't really get the point
02.08.2005 17:41
perfect_radio wrote: PS: A solution was given by Iura in http://www.mathlinks.ro/Forum/viewtopic.php?t=44430 (which is probably correct, but I'm afraid to check it) . I'm wondering whether there is an elegant solution Iura's solution is true, but contain a lot of inessential mistakes. Canonical solution is very short, five lines only.
03.08.2005 00:46
Alexander Khrabrov wrote: perfect_radio wrote: PS: A solution was given by Iura in http://www.mathlinks.ro/Forum/viewtopic.php?t=44430 (which is probably correct, but I'm afraid to check it) . I'm wondering whether there is an elegant solution Iura's solution is true, but contain a lot of inessential mistakes. Canonical solution is very short, five lines only. Glad to hear that Iura's soln is a correct soln. And what do you mean by "canonical"? Anyway, if it is that short, I'd like to see it
03.08.2005 06:21
This problem can be solved by Cauchy's ineq : $ \frac{1}{a^2+\frac{bc}{a}} + \frac{1}{b^2+\frac{ac}{b}} \geq \frac{4}{1+\frac{bc}{a} + \frac{ac}{b}} $ Furthermore , $ \frac{4}{1+\frac{bc}{a} + \frac{ac}{b}} + \frac{bc}{a} + \frac{ac}{b} \geq 1 $ So it's enough to show : $ \frac{c}{c^3+ab} \geq \frac{bc}{a} + \frac{ac}{b} $ The last ineq is true for $ c < c^3 +ab $
03.08.2005 13:09
nttu wrote: $ c < c^3 +ab $ why is this true?
03.08.2005 18:02
If $ a \geq a^3+bc ; b \geq b^3+ac $ and $ c \geq c^3 +bc $ , the ineq is obviously true .
03.08.2005 18:03
nttu wrote: If $ a \geq a^3+bc ; b \geq b^3+ac $ and $ c \geq c^3 +bc $ , the ineq is obviously true . thanks! Your solution is beautiful.
03.08.2005 19:00
At the first line with using Couchy we get $ \frac{1}{a^2+\frac{bc}{a}} + \frac{1}{b^2+\frac{ac}{b}} \geq \frac{4}{a^2+b^2+\frac{bc}{a} + \frac{ac}{b}} $ I do not understand your solution
03.08.2005 19:25
For $p,q,x,y>0$, you can use Cauchy-Schwalz's extended version $\frac{x^2}{p}+\frac{y^2}{q}\geq \frac{(x+y)^2}{p+q}$
03.08.2005 22:50
I know very well this but why $ a^2+b^2=1 $ The problem says $ a^2+b^2+c^2=1 $
03.08.2005 23:07
silouan wrote: I know very well this but why $ a^2+b^2=1 $ The problem says $ a^2+b^2+c^2=1 $ we increase the denominator from $a^2+b^2$ to $a^2+b^2+c^2$. thus, we get smth smaller so the ineq is still true, right?
04.08.2005 16:44
Oh yes! I understand it. i make a thought. We suppose that a<=a^3+bc b<=b^3+ac.............. etc. But If the same time we have that a<=a^3+bc but b>b^3+ac and c>c^3+ab. What happens?
04.08.2005 19:31
silouan wrote: Oh yes! I understand it. i make a thought. We suppose that a<=a^3+bc b<=b^3+ac.............. etc. But If the same time we have that a<=a^3+bc but b>b^3+ac and c>c^3+ab. What happens? No , I don't suppose that . We only need one in three ineqs .
05.08.2005 00:55
nttu wrote: This problem can be solved by Cauchy's ineq : $ \frac{1}{a^2+\frac{bc}{a}} + \frac{1}{b^2+\frac{ac}{b}} \geq \frac{4}{1+\frac{bc}{a} + \frac{ac}{b}} $ Furthermore , $ \frac{4}{1+\frac{bc}{a} + \frac{ac}{b}} + \frac{bc}{a} + \frac{ac}{b} \geq 1 $ So it's enough to show : $ \frac{c}{c^3+ab} \geq \frac{bc}{a} + \frac{ac}{b} $ The last ineq is true for $ c < c^3 +ab $ your solution is correct, and very short also i doubted that such soln exists, but.... Now I have to say it is the most un-natural soln I've seen on ML (there may be other, but i'm not very old on the forum). I mean how on earth did you think of those ineqs
05.08.2005 16:38
Sure! It is a perfect solution but a solution very strange. Has anyone another solution to this problem? Maybe the official?
01.10.2005 00:11
silouan wrote: Sure! It is a perfect solution but a solution very strange. Has anyone another solution to this problem? Maybe the official? It's the official solution. I don't know another solutions except offitial and Iura's one.
03.06.2006 20:34
I have a different solution: 1/(a^2+(bc)/a) + 1/(b^2+(ac)/b) + 1/(c^2+(ba)/a)>=3^2/(a^2+b^2+c^2+ab+bc+ca); the last is equivalent to 2>=ab/c+bc/a+ca/b;then we have to prove that 2abc>=a^2*b^2+c^2*b^2+a^2*c^2 => we have to prove that a^4+b^4+c^4 + 4abc>=1=a^2 + b^2 + c^2 and we have that a^3 + bc>a and we sum with a,b,c and we have what we had to prove.
04.02.2008 13:15
nttu wrote: $ \frac {c}{c^3 + ab} \geq \frac {bc}{a} + \frac {ac}{b}$ The last ineq is true for $ c < c^3 + ab$ Why is $ \frac {c}{c^3 + ab} \geq \frac {bc}{a} + \frac {ac}{b}$ true for $ c < c^3 + ab$?
04.02.2008 14:49
Clear the denominators and you'll see why
29.05.2010 08:17
nttu wrote: If $ a \geq a^3+bc ; b \geq b^3+ac $ and $ c \geq c^3 +bc $ , the ineq is obviously true . your solution is not true!
17.11.2015 03:07
http://www.artofproblemsolving.com/community/c6h44430p5551371 http://www.artofproblemsolving.com/community/c6h46087p297181