work 004

geniusbliss

Prove that for positive reals $a,b,c$ with sum 2 ,
$\frac {a}{b(a + b)} + \frac {b}{c(b + c)} + \frac {c}{a(a + c)} \ge 2$

**Posted:** Sun Nov 08, 2009 4:06 am

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Dimitris X

$LHS\ge \frac {4}{ \sum_{sym} a^2b}$ .

So we only need to prove that:
$2\ge \sum_{sym}a^2b \Longleftrightarrow (a+b+c)^3 \ge 4(\sum_{sym}a^2b)$
which is true clearly true.

**Posted:** Sun Nov 08, 2009 4:45 am

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ΠΑΙΡΝΩ ΤΑΜΠΕΛΑ ΚΑΙ ΕΓΩ ΤΟΥ ΕΘΝΙΚΟΥ ΠΡΟΔΟΤΗ ΑΦΙΕΡΩΜΕΝΟ ΚΑΙ ΑΥΤΟ ΣΕ ΚΑΘΕ ΔΟΥΛΟ ΠΑΤΡΙΩΤΗ.....

geniusbliss

ok..
By AM-GM,
$\sum_{cyclic}\frac {a}{b(a + b)} + (a + b)a + ab \ge 3a$
which is equicalent to
$\sum_{cyclic}\frac {a}{b(a + b)} \ge 6 - 4 = 2$
the inequality also holds for the condition $a+b+c=1$

**Posted:** Sun Nov 08, 2009 5:32 am

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Pain rinnegan · Poincare Conjecture Offline Joined: 16 Apr 2009 Posts: 165

Equality can't occur .

**Posted:** Sun Nov 08, 2009 5:47 am

geniusbliss

**Posted:** Sun Nov 08, 2009 8:16 pm

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Quis custodiet ipsos custodes
Mathematical Dreams Mr. Green