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#26 2012-07-09 22:22:48
- hcj73jx
- Member
- Registered: 2012-02-29
- Posts: 12
Re: Help me!!!a>0,b>0,c>0,a+b+c=1,0<p<1,prove that:
a+b+c=1,prove \; that\sum \frac{\sqrt{a}}{1-a}\geqslant \frac{3\sqrt{3}}{2};prove:\left ( a+b+c \right )^{3}=\left ( \sum \sqrt[3]{\frac{\sqrt{a}}{1-a}} \ast \sqrt[3]{\frac{\sqrt{a}}{1-a}} \ast \sqrt[3]{{a^{2}(1-a)^{2}}} \right )^{3}
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#27 2012-07-09 22:31:03
- bobbym
- bumpkin
- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606
Re: Help me!!!a>0,b>0,c>0,a+b+c=1,0<p<1,prove that:
I latexed it for you. To latex it yourself you must put it between math tags.
prove that
prove:
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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#28 2012-07-10 00:03:55
- hcj73jx
- Member
- Registered: 2012-02-29
- Posts: 12
Re: Help me!!!a>0,b>0,c>0,a+b+c=1,0<p<1,prove that:
Last edited by hcj73jx (2012-07-10 01:00:08)
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#29 2012-07-10 02:24:06
- bobbym
- bumpkin
- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606
Re: Help me!!!a>0,b>0,c>0,a+b+c=1,0<p<1,prove that:
Hi;
Forgive me but is that your proof or more questions?
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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