nfornick Posted September 22, 2005 Share Posted September 22, 2005 Show that a/c + b/a + c/b > a+b+c, for a, b, c > 0 and a*b*c<1 Link to comment Share on other sites More sharing options...
BigMoosie Posted September 22, 2005 Share Posted September 22, 2005 [math]\frac{a}{c} + \frac{b}{a} + \frac{c}{b} > a+b+c[/math] [math]a + b + c > (a+b+c)(abc)[/math] [math]0 > (a+b+c)(abc - 1)[/math] Link to comment Share on other sites More sharing options...
timo Posted September 22, 2005 Share Posted September 22, 2005 That´s not correct. The 2nd line would be: a²b + b²c + c²a > (a+b+c)(abc) Link to comment Share on other sites More sharing options...
albertlee Posted September 22, 2005 Share Posted September 22, 2005 extending from Atheist, a2b + b2c + c2a > (a+b+c)(abc) ==> a2b + b2c + c2a > a2bc+ab2c+abc2 therefore: 0 > (c-1)a2b+(a-1)b2c+(b-1)ac2 Link to comment Share on other sites More sharing options...
Mobius Posted September 22, 2005 Share Posted September 22, 2005 (c-1)a2b+(a-1)b2c+(b-1)ac2 And how are you proving that this is a negative number? I don't seem to be ble to solve this inequality mathematically BUT I can solve it logically. In order to satisfy a*b*c<0 where they are all positive values then at least ONE of them must be a fraction and the other two numbers cannot multiply together to be greater than the denominator. e.g. 1/20, 3, 5... When multiplied this gives 3/4. Now:- a/c + b/a + c/b > a+b+c On the left hand side a, b and c are all denominators, one of them will be a fraction (say a) and as it is on the bottom line it will be inverted and will be large. It will also be larger then the other two numbers added together (b + c) due to my logic above (a*b*c<0). i.e. if it is larger than the product it will be larger than the sum. So the right hand side will be b + c plus the fraction which will be smaller than the left had side. Of course if a, b and c are all fractions then this inequality is obvious as when fractions are divided the answer is always greater (or equal) then either of the two fractions to begin with. this is not very mathematical and I'm sure if there is a flaw in my logic it will be pointed out.... It is a logical proof however (I did try for a mathematical one!). Link to comment Share on other sites More sharing options...
albertlee Posted September 22, 2005 Share Posted September 22, 2005 let me rephrase... Link to comment Share on other sites More sharing options...
Mobius Posted September 23, 2005 Share Posted September 23, 2005 Your last equation 0 > (c-1)a^2b+(a-1)b^2c+(b-1)ac^2, would make sense if a, b and c were ALL fractions, but as I pointed out in my last post only 1 of them has to be a fraction (according to the original conditions). I think my proof is fine but it may have a flaw... Awaiting confirmation!!!! Link to comment Share on other sites More sharing options...
BigMoosie Posted September 23, 2005 Share Posted September 23, 2005 That´s not correct. The 2nd line would be: a²b + b²c + c²a > (a+b+c)(abc) oops, stupid me Link to comment Share on other sites More sharing options...
nfornick Posted September 24, 2005 Author Share Posted September 24, 2005 Re: Mobius a*b*c<1 but not a*b*c<0 "at least ONE of them must be a fraction" means one or two may be a fraction "one of them will be a fraction (say a) and as it is on the bottom line it will be inverted and will be large. It will also be larger then the other two numbers added together (b + c)" a*b*c<1 does not imply 1/a>b+c say a=1/3, b=1, c=2 Link to comment Share on other sites More sharing options...
Mobius Posted September 24, 2005 Share Posted September 24, 2005 "at least ONE of them must be a fraction" means one or two may be a fraction Doesn't matter of 2 of them are fractions, this just strengtens the case. 1/a>b+c say a=1/3, b=1, c=2 That is true, in this case 1/a = b + c However there are two more numbers to be added to that 1/a (or b/a) as it would be. Now I did say it would be greater but as you showed it could be equal, however as there are 2 more numbers to be added to it, it will always be greater... Link to comment Share on other sites More sharing options...
dttom Posted September 28, 2005 Share Posted September 28, 2005 L.H.S. a/c + b/a + c/b =[(a^2)b+(b^2)c+(c^2)a]/abc let x as abc which is smaller than 1 =(a^2b+b^2c+c^2a)/x L.H.S.*x=a^2b+b^2c+c^2a R.H.S*x=(a+b+c)abc =(a^2)bc+a(b^2)c+ab(c^2) =(c-1)(a^2b)+(a-1)(b^2c)+(b-1)(c^2a)+a^2b+b^2c+c^2a let LHS*x as r and RHS*x as w, r-r=0 w-r=(c-1)(a^2b)+(a-1)(b^2c)+(b-1)(c^2a) 0>abc(a+b+c)-(a^2b+b^2c+c^2a) Link to comment Share on other sites More sharing options...
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