again a+b+c=1

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Corei13
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again a+b+c=1

Unread post by Corei13 » Tue Dec 21, 2010 11:18 pm

Prove that, for $a+b+c=1$,
\[ \sum_{a,b,c}{\left(\frac{2}{a^{2}}+\frac{7}{a}\right)}\ge 9+\frac{4}{abc} \]
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ishfaqhaque
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Re: again a+b+c=1

Unread post by ishfaqhaque » Sun Jan 16, 2011 11:19 pm

homogenous policy should work here.....

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Moon
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Re: again a+b+c=1

Unread post by Moon » Sun Jan 16, 2011 11:33 pm

I don't think that it is homogeneous. Probably you meant that we need to homogenize. However, I don't think that it will be an easy job!
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ishfaqhaque
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Re: again a+b+c=1

Unread post by ishfaqhaque » Mon Jan 17, 2011 2:12 pm

well, we used to use some stupid brute forces for several inequalities .....i told this from that experience.

but it should not be so hard to homogenize everything

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