S.O.S. Mathematics CyberBoard

Your Resource for mathematics help on the web!
It is currently Thu, 20 Jun 2013 05:18:30 UTC

View unanswered posts | View active topics


Board index » High School and College Mathematics » Algebra

All times are UTC [ DST ]



Post new topic Reply to topic  Page 1 of 1
  [ 4 posts ] 
  Print view Previous topic | Next topic 
Author Message
kazuya
 Post subject: please help me
PostPosted: Thu, 3 May 2012 19:25:00 UTC 
Offline
Math Cadet

Joined: Wed, 22 Nov 2006 14:07:10 UTC
Posts: 5
when a,b,c is positive real numbers
how to proof

1/a(1+b) + 1/b(1+c) +1/c(1+a) >= 3/(1+abc)
Top
 Profile  
 
Denis
 Post subject: Re: please help me
PostPosted: Thu, 3 May 2012 20:43:42 UTC 
Offline
Member of the 'S.O.S. Math' Hall of Fame

Joined: Sun, 24 Jul 2005 20:12:39 UTC
Posts: 3724
Location: Ottawa Ontario
kazuya wrote:
when a,b,c is positive real numbers
how to proof
1/a(1+b) + 1/b(1+c) +1/c(1+a) >= 3/(1+abc)

Denominators? Do you mean 1/(a(1+b)) or (1+b)/a ?

_________________
Walked over to a beggar...he gave me a quarter...
Top
 Profile  
 
kazuya
 Post subject: Re: please help me
PostPosted: Fri, 4 May 2012 01:17:35 UTC 
Offline
Math Cadet

Joined: Wed, 22 Nov 2006 14:07:10 UTC
Posts: 5
Denis wrote:
kazuya wrote:
when a,b,c is positive real numbers
how to proof
1/a(1+b) + 1/b(1+c) +1/c(1+a) >= 3/(1+abc)

Denominators? Do you mean 1/(a(1+b)) or (1+b)/a ?


1/(a(1+b))，others the same，thanks a lot
Top
 Profile  
 
rdj5933mile5math64
 Post subject: Re: please help me
PostPosted: Sat, 5 May 2012 17:01:21 UTC 
Offline
Senior Member
User avatar

Joined: Wed, 4 Apr 2012 03:51:40 UTC
Posts: 129
Location: Hockeytown aka Detroit
Hmmm I feel like there's probably a more clever way of proving this.

Note the way I presented this solution sketch is not the way one should present a normal solution, however, it shows how one should do this problem.

Hint:
Spoiler:
You have variables in the denominator. What can you do to eliminate this problem?



Answer to Hint:
Spoiler:
Multiply both expressions by abc(1+a)(1+b)(1+c)(1+abc).



By doing the stuff in Answer:
Spoiler:
Let RHS be the Right Hand Side expression and LHS be the Left Hand Side expression.

Note that by multiplying both sides by abc(1+a)(1+b)(1+c)(1+abc).
(b(1+c)c(1+a)+c(1+a)a(1+b)+a(1+b)b(1+c))(1+abc)=RHS
3abc(1+a)(1+b)(1+c)=LHS

Expanding stuff yields:
(bc+abc+bc^2+abc^2+ac+abc+a^2c+a^2bc+ab+abc+ab^2+ab^2c)(1+abc)=RHS
3abc(1+a+b+c+ab+bc+ca+abc)=LHS

By collecting a few similar terms in the RHS yields,
((ab+bc+ca)+3abc+(bc^2+a^2c+ab^2)+(abc)(a+b+c))(1+abc)=RHS
3abc(1+a+b+c+ab+bc+ca+abc)=LHS

We now multiply out the (1+abc) in the RHS.
(ab+bc+ca)+3abc+(bc^2+a^2c+a^2b)+(abc)(a+b+c)+abc(ab+bc+ca)+ 3(abc)^2+(abc)(bc^2+a^2c+ab^2)+(abc)^2(a+b+c)=RHS
3abc(1+a+b+c+ab+bc+ca+abc)=LHS

We subtract 3abc+ 3(abc)^2+(abc)(a+b+c)+abc(ab+bc+ca) from both the RHS and the LHS to get:
(ab+bc+ca)+(bc^2+a^2c+ab^2)+ (abc)(bc^2+a^2c+ab^2)+(abc)^2(a+b+c)=RHS
2abc(a+b+c)+2abc(ab+bc+ca)=LHS


Well this looks a lot easier than what we had before! How do we finish the problem off?
Spoiler:
Note that by AM-GM:
ca+(abc)(bc^2) \ge 2abc^2
ab+(abc)(a^2c) \ge 2a^2bc
bc+(abc)(ab^2) \ge 2ab^2c

Adding these three inequalities yields:
(ab+bc+ca)+ (abc)(bc^2+a^2c+ab^2) \ge 2abc(a+b+c) (X)

Similarly, we note that once again by AM-GM:
(abc)^2(a)+ab^2 \ge 2abc(ab)
(abc)^2(b)+bc^2 \ge 2abc(bc)
(abc)^2(c)+ca^2 \ge 2abc(ca)

Adding these three inequalities yields:
(bc^2+a^2c+ab^2)+(abc)^2(a+b+c) \ge 2abc(ab+bc+ca) (Y)

Adding inequalities (X) and (Y) gives us the desired inequality.

_________________
math puns are the first sine of madness
-JDR
:mrgreen:
Top
 Profile  
 
Display posts from previous:  Sort by  
Post new topic Reply to topic  Page 1 of 1
  [ 4 posts ] 

Board index » High School and College Mathematics » Algebra

All times are UTC [ DST ]

Who is online

Users browsing this forum: No registered users

You cannot post new topics in this forum
You cannot reply to topics in this forum
You cannot edit your posts in this forum
You cannot delete your posts in this forum

Search for:
Jump to:  
Contact Us | S.O.S. Mathematics Homepage
Privacy Statement | Search the "old" CyberBoard

users online during the last hour
Powered by phpBB © 2001, 2005-2011 phpBB Group.
Copyright © 1999-2013 MathMedics, LLC. All rights reserved.
Math Medics, LLC. - P.O. Box 12395 - El Paso TX 79913 - USA