Prove that, for $a+b+c=1$,
\[ \sum_{a,b,c}{\left(\frac{2}{a^{2}}+\frac{7}{a}\right)}\ge 9+\frac{4}{abc} \]
again a+b+c=1
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Re: again a+b+c=1
homogenous policy should work here.....
Re: again a+b+c=1
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!
"Inspiration is needed in geometry, just as much as in poetry." -- Aleksandr Pushkin
Please install LaTeX fonts in your PC for better looking equations,
learn how to write equations, and don't forget to read Forum Guide and Rules.
Please install LaTeX fonts in your PC for better looking equations,
learn how to write equations, and don't forget to read Forum Guide and Rules.
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- Posts:20
- Joined:Thu Dec 09, 2010 3:30 pm
Re: again a+b+c=1
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
but it should not be so hard to homogenize everything