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Prove this inequality 
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#1
Nov2012, 08:12 AM

PF Gold
P: 833

1. The problem statement, all variables and given/known data
If a,b,c are the positive real numbers, prove that [itex]a^2(1+b^2)+b^2(1+c^2)+c^2(1+a^2) \geq 6abc[/itex] 2. Relevant equations 3. The attempt at a solution With a little simplification L.H.S = [itex](a^2+b^2+c^2)+(a^2b^2+b^2c^2+c^2a^2)[/itex] Using A.M>=G.M [itex]\dfrac{a^2+b^2+c^2}{3} \geq (a^2b^2c^2)^{\frac{1}{3}} \\ a^2+b^2+c^2 \geq 3a^{2/3}b^{2/3}c^{2/3} \\ [/itex] Also [itex] \dfrac{a^2b^2+b^2c^2+c^2a^2}{3} \geq (a^2b^2.b^2c^2.c^2a^2)^{1/3} \\ a^2b^2+b^2c^2+c^2a^2 \geq 3a^{4/3}b^{4/3}c^{4/3} [/itex] Adding the two inequalities [itex] (a^2+b^2+c^2)+(a^2b^2+b^2c^2+c^2a^2) \geq 3[a^{2/3}b^{2/3}c^{2/3}+a^{4/3}b^{4/3}c^{4/3}] [/itex] Now how do I simplify next? 


#2
Nov2012, 08:24 AM

Mentor
P: 11,831

Your last inequality can be written as
LHS >= 3 (x+x^2) with an appropriate x. And you have to show that LHS >= 6x^{3/2} You can just use the same trick again at your new sum. 


#3
Nov2012, 08:38 AM

PF Gold
P: 833




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