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I hope I'm not spamming because I have alot to learn still....
But anyways have you tried
if abc= 1 then 1 + 3 ≥ 6
___ ____
Variable Double Variable
I'm thinking that A+B+C = 1:D plz dont get mad if it seems off the wall.
and it may go the same with the other one but with two abc variable but mixed up which adds to 2, so it might look like this.....
1+ 3 ≥ 6
--- --- = 4 ≥ 6
1 2 --- --- = 4 ≥ 3 to me it seems to prove the problem hope i helped
1 2
Last edited by Kurtz (2008-08-08 14:07:29)
I am a mathemagician. You ask why? I point up
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I'm thinking that A+B+C = 1
a, b, and c must all be positive. Thus, if a+b+c = 1, then they must all be less than 1 and greater than 0. However, then it has to be a*b*c < 1, which contradicts the fact that a*b*c = 1.
"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."
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So what if A*B*C=1 then wouldnt it also all be 1?
because A times B times C = 1 which could be substituted as A=1 B=1 C=1
which is A(1)*B(1)*C(1)=1
Then we're proving the equation false?
I am a mathemagician. You ask why? I point up
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Kurtz, try A = 1/2, B = 2, C = 1
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It appears with number fiddling that:
(ab+bc+ca-3)*2 > a+b+c-3 when ab+bc+ca < 6 anyway.
For ab+bc+ca > 6, it doesn't matter anymore.
Last edited by John E. Franklin (2008-08-08 18:06:28)
igloo myrtilles fourmis
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perhaps the solution lies in the relationship with those equations in a,b,c and a cubic polynomial:
a b c are roots of the cubic equation x³ + Ax² + Bx + C = 0 such that.
abc = -A = 1
ab+bc+ac = B
a+b+c = -C
also ofcourse; x³ + Ax + Bx + C = (x-a)(x-b)(x-c)
and so far we have: x³ - x² + Bx + C = 0
since a,b,c are positive, C must be negative, and B must be positive, hence;
well now i'm stuck and tired and going to bed ;P
The Beginning Of All Things To End.
The End Of All Things To Come.
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a b c are roots of the cubic equation x³ + Ax² + Bx + C = 0 such that.
abc = -A = 1
ab+bc+ac = B
a+b+c = -C
You got the A and the C in the wrong places.
Anyway, while floundering in the sea of despair, I stumbled upon another inequality:
My attempt was as follows. If
, we have equality. Otherwise, given that , we must have either or .I can prove the inequality for the case
.But I cant do it for the case
.Ooh, I was sooo close!
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Hopefully this works.
Let
. We have .So the inequality is
By CauchySchwarz
.Hence, we have
and we are done if we can show that
.Well, writing
, we haveand we are indeed done.
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Who said that you're USELESS......?
If two or more thoughts intersect, there has to be a point!
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uhm didnt work out
(why do sometimes this forum post when i click preview??)
Last edited by Kurre (2008-08-09 21:47:19)
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What didnt work out?
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Well I was writing a solution and when I clicked preview It got posted instead and then I didnt manage to complete the proof. I substituted c=1/ab and the inequality for your case reduced to:
Last edited by Kurre (2008-08-09 23:58:59)
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I tried that too. In fact,
and so I tried using the result I had already proved:It was no good.
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tony123 strikes.
http://www.mathlinks.ro/viewtopic.php?p=1219810#1219810
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Set
. With some calculations, we get the new problem accompanying the condition
By AM-GM, we have the simple inequality: in order to imply the important result
( which is what we want! )
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The inequality is equivalent to
By AM-GM Inequality,
It is obvious that
,so we are done!
Last edited by tony123 (2008-08-11 00:06:43)
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The inequality is equivalent to
By AM-GM Inequality,
.It is obvious that
,so we are done!
Thank you!
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oh I though a,b,c were = 1 so that:
1+3/((1)+(1)+(1)) 1+6/((1*1)+(1*1)+(1*1))
1+1=2 1+2=3. Oh but then 2 is not equal to 3
Hmm... So who figured it out?
π≈ 3.141 592 653 589 793 238 462 643 383 279 502 884 197 169 399 375 105 820 974 944 592 307 816 406 286 208 998 628 034 825 342 117 067 982 148 086 513 282 306 647 093 844 609 550 582 231 725 359 408 128 481 117 450 284 102 701 938 521 105 559 644 622 948 954 930...
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Moose that thinking process has been done. I believe I post it above but also another thing is that (A)(B)(C) can not all equal 1 and like Identity showed me is that we must use invert substitution that will also give us the result of one. Not just 1*1*1.
I am a mathemagician. You ask why? I point up
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