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#1 2012-07-02 20:35:27
- hcj73jx
- Member
- Registered: 2012-02-29
- Posts: 12
Help me!!!a>0,b>0,c>0,a+b+c=1,0<p<1,prove that:
Help me!!!a>0,b>0,c>0,a+b+c=1,0<p<1,prove that:
a^p/(1-a)+b^p/(1-b)+c^p/(1-c)≥3^(2-p)/2
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#2 2012-07-02 20:43:12
- 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;
Welcome to the forum. Is this what the problem looks like?
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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#3 2012-07-03 01:29:55
- 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:
\frac{a^p}{1-a}+\frac{b^p}{1-b}+\frac{c^p}{1-c}\geq \frac{3^(2-p)}{2}
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#4 2012-07-03 01:41:06
- 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;
You just copied what I wrote. Is post #2 what you want to 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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#5 2012-07-03 01:44:37
- anonimnystefy
- Real Member
- From: Harlan's World
- Registered: 2011-05-23
- Posts: 16,049
Re: Help me!!!a>0,b>0,c>0,a+b+c=1,0<p<1,prove that:
Hi bobbym
They are not the same. Notice that the last 2 is in different places in your and her posts.
Here lies the reader who will never open this book. He is forever dead.
Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment
The knowledge of some things as a function of age is a delta function.
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#6 2012-07-03 01:46:25
- 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:
When I latex the new one it does not make a whole lot of sense.
A linear function on the RHS?
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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#7 2012-07-03 01:51:09
- 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:
Hi;
\frac{3^(2-p)}{2}
Not
3^{\frac{(2-p)}{2}}
thank you!
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#8 2012-07-03 01:53:55
- 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:
This is what you have:
it does not make sense because it is incorrectly latexed.
Why not use this site here for all your latexing?
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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#9 2012-07-03 02:02:02
- 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:
\frac{3^{(2-p)}}{2}
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#10 2012-07-03 02:04:15
- 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:
So this is it?
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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#11 2012-07-03 02:16:01
- 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:
yes,thank you!
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#12 2012-07-03 02:17:01
- 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 hcj73jx;
Okay that is what I wrote.
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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#13 2012-07-03 02:31:12
- 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:
yes,thank you!
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#14 2012-07-03 02:37:53
- 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;
Okay, I will work on it.
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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#15 2012-07-03 08:53:40
- TheDude
- Member
- Registered: 2007-10-23
- Posts: 361
Re: Help me!!!a>0,b>0,c>0,a+b+c=1,0<p<1,prove that:
First show that
is minimal when a = b = c = 1/3. From there:
Last edited by TheDude (2012-07-03 08:54:15)
Wrap it in bacon
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#16 2012-07-03 09:01:44
- zetafunc.
- Guest
Re: Help me!!!a>0,b>0,c>0,a+b+c=1,0<p<1,prove that:
First show that
is minimal when a = b = c = 1/3.
How do you do this, out of interest?
#17 2012-07-04 20:54:13
- 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:
How do you do it?help me.
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#18 2012-07-04 22:39:06
- 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 hcj73jx;
No one has been able to just yet.
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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#19 2012-07-06 04:46:15
- TheDude
- Member
- Registered: 2007-10-23
- Posts: 361
Re: Help me!!!a>0,b>0,c>0,a+b+c=1,0<p<1,prove that:
Here's what I've got so far. Rewrite the LHS to read
The partial derivative with respect to a is
Setting this equal to 0 gives us
You can show from here that a = b = c = 1/3 is a critical point, but I haven't been able to show that it's the only critical point.
Wrap it in bacon
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#20 2012-07-06 08:39:47
- TheDude
- Member
- Registered: 2007-10-23
- Posts: 361
Re: Help me!!!a>0,b>0,c>0,a+b+c=1,0<p<1,prove that:
Unless I made a mistake somewhere I believe that the inequality is wrong. If you let a be a very small number (like 10^-10), b = 0.5, and c = 1 - a - b, and p = 0.25 the inequality will fail.
Wrap it in bacon
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#21 2012-07-06 08:54:24
- 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 TheDude;
Yes, I agree. That is quite common.
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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#22 2012-07-09 01:25:07
- 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:
Thank you.
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#23 2012-07-09 01:27:46
- 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 hcj73jx;
That is what you wanted?!
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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#24 2012-07-09 14:13:18
- 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:
Hi，Thank you，I have proved p=1/2 the inequality is right.
I want to know p=1/3 the inequality is right too? how prove? pln3/2>ln9/8 the inequality is right too?right?
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#25 2012-07-09 21:02:19
- 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 hcj73jx;
How did you prove it for p = 1 / 2 ?
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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