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kazuya
 Post subject: please help me
PostPosted: Thu, 3 May 2012 19:25:00 UTC 
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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)
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Denis
 Post subject: Re: please help me
PostPosted: Thu, 3 May 2012 20:43:42 UTC 
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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 ?

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kazuya
 Post subject: Re: please help me
PostPosted: Fri, 4 May 2012 01:17:35 UTC 
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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
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rdj5933mile5math64
 Post subject: Re: please help me
PostPosted: Sat, 5 May 2012 17:01:21 UTC 
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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.

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