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Hi;
This one is pretty tough but see what you can do.
For positive numbers a,b,c:
Last edited by bobbym (2009-09-03 09:27:24)
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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Hi bobbym
I am online now just to say hi
Riad Zaidan
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Last edited by JaneFairfax (2009-09-03 22:11:32)
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Hi Jane;
Nice solution!
Last edited by bobbym (2009-09-04 15:21:51)
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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[align=center]
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Nobody?
Please give it a try. Its actually very simple.
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Please give it a try. Its actually very simple.
Please forgive me Jane, I have been working on it. I think it is one you give to your worst enemy.
Last edited by bobbym (2009-09-11 06:56:00)
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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Hi Jane;
I couldn't get it, can you show me your solution. You know, the simple one. I have exhausted all of my ideas.
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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Why did the vector cross the road?
It wanted to be normal.
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Hi mathsyperson;
I avoided the idea of minima because I don't think that is the way Jane went. Now after lots of dead ends your hint is my only chance.
Prove:
Minimizing the first radical of A:
Minimum at b = 2a
Minimizing the second radical of A:
minimum at b = 2c
Plugging a = b / 2 and c = b /2 into:
We get:
Squaring both sides to make use of the fact that:
Since the 3 b^2 on the left represents the minimum for the LHS of A) we have proved:
Last edited by bobbym (2009-09-12 17:20:55)
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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If you want to avoid calculus, you could also say
Clearly the squared part is non-negative and so the minimal value must be -b²/4.
Why did the vector cross the road?
It wanted to be normal.
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Last edited by JaneFairfax (2009-09-13 14:38:23)
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Hi Jane;
Thanks for your method and thanks mathsy for yours.
Last edited by bobbym (2009-09-13 14:27:15)
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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A problem similar to mine can be found here.
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Hi Jane;
Thanks so much for the link. I appreciate that. Hmmm, that ValentineA again ?!
Last edited by bobbym (2009-09-15 08:08:17)
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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