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Ask iit jee aieee pet cbse icse state board community Community Discussion Question: easy --
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[Post New]26 Mar 2008 19:00:31 IST
    Subject: easy --
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anchitsaini (4352)

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show that for all (positive) real numbers a,b,c

[ 3  ] [ 4a3 +  4b3 ] + [ 3  ] [ 4b3 +  4c3 ] +[ 3  ]   [ 4c3 +  4a3 ]

<=

4a2 / [a+b ]   +   4b2 / [b+c]   +   4c2 / [a+c]
Impossible To be Impossible is Impossible
    
[Post New]26 Mar 2008 19:03:31 IST
    Subject: Re:easy --
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Werewolf (333)

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ok bubba boy go ahead
"All of us are God's creatures... just some are more creature than others."
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[Post New]26 Mar 2008 19:07:30 IST
    Subject: easy --
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anchitsaini (4352)

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posted
Impossible To be Impossible is Impossible
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[Post New]26 Mar 2008 19:46:01 IST
    Subject: easy --
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anchitsaini (4352)

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my mistake
it is all positive real numbers
not for all real numbers
Impossible To be Impossible is Impossible
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[Post New]26 Mar 2008 20:31:29 IST
    Subject: Re:easy --
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sboosy (3063)

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\mbox{Using AM of mth power <= mth power of AM for m<1 , we get} \\ \\ \frac{(4a^3+4b^3)^{\frac{1}{3}}+(4b^3+4c^3)^{\frac{1}{3}}+(4c^3+4a^3)^{\frac{1}{3}}}{3} <= (\frac{8(a^3+b^3+c^3)}{3})^{\frac{1}{3}} \\ \\ \frac{a^3+b^3+c^3}{3} >= abc \\ \\ \frac{(4a^3+4b^3)^{\frac{1}{3}}+(4b^3+4c^3)^{\frac{1}{3}}+(4c^3+4a^3)^{\frac{1}{3}}}{3} <= 6(abc)^\frac{1}{3} ......(1) \\ \\ \\ \mbox{Using lemma} \ \frac{x^2}{a}+\frac{y^2}{b} >= \frac{(x+y)^2}{a+b} \\ \\ \mbox{We get} \\ \\ \frac{4a^2}{a+b}+\frac{4b^2}{b+c}+\frac{4c^2}{a+c}>=2(a+b+c)>=6(abc)^\frac{1}{3} .....(2) \\ \\ \mbox{Comparing 1 and 2 we get the result} \\ \\ \mbox{Thus we have} \\ \\ \frac{(4a^3+4b^3)^{\frac{1}{3}}+(4b^3+4c^3)^{\frac{1}{3}}+(4c^3+4a^3)^{\frac{1}{3}}}{3} <=6(abc)^\frac{1}{3}<= \frac{4a^2}{a+b}+\frac{4b^2}{b+c}+\frac{4c^2}{a+c}
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[Post New]26 Mar 2008 20:33:08 IST
    Subject: easy --
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sboosy (3063)

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Correction first line
AM of mth power <= mth power of AM for m<1
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[Post New]26 Mar 2008 20:54:45 IST
    Subject: Re:easy --
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anchitsaini (4352)

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sboosy deserves a round of applause

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