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Inequality problem trouble

  1. Oct 26, 2005 #1
    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 :smile:
     
  2. jcsd
  3. Oct 26, 2005 #2

    mathman

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    Set the largest number aside (it must be greater than 1, since otherwise any product of 11 will be <1). You have now 9 distinct sets of 11, all products >1, therefore product of 99 >1, multiply in the set aside number and you're done.
     
  4. Oct 26, 2005 #3
    OK, great thanks! It's clear now! :smile:

    So in particular it means that we have at least 10, and not 9, numbers greater than 1, doesn't it?
    I'm arguing with my friend about that...:grumpy:...because he insists on 9!

    How can we show mathematically, I mean more formally, that 10 is right...that the largest number is greater than 1? :smile:
     
  5. Oct 26, 2005 #4

    shmoe

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    Far more than 10 of them are greater than 1. Any set of 11 will have a number greater than 1, no matter how you choose this subset. If you had 90 numbers less than 1, you'd have plenty of ways to pick a subset of 11 whose product is less than 1.

    Try to answer from this direction: what's the most that you can have less than (or equal to) 1?
     
  6. Oct 28, 2005 #5
    Is it right if I say 50/50?...I mean if 51 numbers are <1 than the whole product will be <1?
     
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