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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)

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Joined: Sun, 24 Jul 2005 20:12:39 UTC
Posts: 3692
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 ?

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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

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 Senior Member

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?

Spoiler:
Multiply both expressions by .

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 .

Expanding stuff yields:

By collecting a few similar terms in the RHS yields,

We now multiply out the in the RHS.

We subtract from both the RHS and the LHS to get:

Well this looks a lot easier than what we had before! How do we finish the problem off?
Spoiler:
Note that by AM-GM:

(X)

Similarly, we note that once again by AM-GM:

(Y)

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

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