Manifold Posted October 26, 2005 Share Posted October 26, 2005 Hi, I've just picked up the following problem and got stuck...although seems easy. You have 100 positive real numbers. The product of each 11 of them (11 different) is greater than 1. Show that the product of all hundred numbers is greater than 1. Please, give me a hint first. Thanks Link to comment Share on other sites More sharing options...

timo Posted October 27, 2005 Share Posted October 27, 2005 If I understood you correctly, then the product of the 11 smallest numbers is >1. Link to comment Share on other sites More sharing options...

cosine Posted October 27, 2005 Share Posted October 27, 2005 If I understood you correctly, then the product of the 11 smallest numbers is >1. Hm, yes, this is neccessary, given the hypothesis! Awesome observation! Given that, I believe a proof could follow this way... I don't know if the original poster wanted to see one though, so I present a spoiler space: [hide]Since any collection of 11 numbers is > 1, then let A = the product of the smallest 11 numbers. By hypothesis, A > 1. Let B = the product of all the other numbers of the collection. Then the question asks to proove if AB > 1. We know that A > 1. Since A is the smallest 11 members of the set, it follows that B > A, so B > 1. Thus, AB > 1. This being what we wanted to proove.[/hide] Link to comment Share on other sites More sharing options...

shmoe Posted October 27, 2005 Share Posted October 27, 2005 [hide]Since A is the smallest 11 members of the set' date=' it follows that B > A... [/hide'] This doesn't follow without more work: [hide]Namely the product of the 11 smallest numbers in a set isn't necessarily less than the product of the rest of the numbers. Since you have 89>=11 other numbers, it will be sufficient to show that these are all greater than 1 (it would also be enough to show the 78 largest are >=1). This follows since at least one of the smallest 11 must be greater than 1, else we couldn't have A>1[/hide] Link to comment Share on other sites More sharing options...

cosine Posted October 27, 2005 Share Posted October 27, 2005 This doesn't follow without more work: [hide]Namely the product of the 11 smallest numbers in a set isn't necessarily less than the product of the rest of the numbers. Since you have 89>=11 other numbers' date=' it will be sufficient to show that these are all greater than 1 (it would also be enough to show the 78 largest are >=1). This follows since at least one of the smallest 11 must be greater than 1, else we couldn't have A>1[/hide'] Thankyou for strengthening the rigor Link to comment Share on other sites More sharing options...

Manifold Posted October 28, 2005 Author Share Posted October 28, 2005 Well, I understood why we took the 11 smallest numbers, but I don't quite understood what is actually meant by this. The 11 smallest of what...of which sets/subsets? Or do you simply mean that in the product of 100 numbers one can pick 11 whose product is fewer than the the product of the rest?...which follows from: if ab>1 then either a>1 and b>1 or a>1 and b<1 or a<1 and b>1...? Aren't the expressions in terms of this and in terms of "smallest" self-excluding? (Although if so, this doesn't make the solution wrong!) Link to comment Share on other sites More sharing options...

timo Posted October 28, 2005 Share Posted October 28, 2005 Well, I understood why we took the 11 smallest numbers, but I don't quite understood what is actually meant by this. The 11 smallest of what...of which sets/subsets? I don´t really understand your problem. You are given 100 positive real numbers and take the 11 smallest ones. Or do you simply mean that in the product of 100 numbers one can pick 11 whose product is fewer than the the product of the rest? No, I meant the set of 11 of those hundred numbers which gives the smallest product of all possible sets of 11 of those hundred numbers (because they are the smallest numbers). ...which follows from: if ab>1 then either a>1 and b>1 or a>1 and b<1 or a<1 and b>1...? Aren't the expressions in terms of this and in terms of "smallest" self-excluding? (Although if so, this doesn't make the solution wrong!) Sorry, I didn´t understand that part at all. Link to comment Share on other sites More sharing options...

Manifold Posted October 29, 2005 Author Share Posted October 29, 2005 OK, forget the last thing you don't understand ...I now see myself that it's a bit of nonsense... I got what you meant and have a clear picture of the matter now. Thanks a lot! Link to comment Share on other sites More sharing options...

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