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Given a set $S$ of ten positive real numbers whose product is $32$, show that $S$ contains six numbers whose product is at least $8$?

I tried to prove it, but it seems that the question is ambiguous. Say, consider a set of six numbers. If their product is $1$, then the remaining $4$ numbers should have product of $32 (4\times 2\times 2\times 2)$. So, there are six numbers whose product is $1$, less than $8$. This contradicts the question.

anand konjengbam
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  • Welcome to Math.SE! Please add more context; what have you tried, what is the motivation for studying this problem etc. Also, let me tell you that on Math.SE it is considered rude to use imperatives like "prove this", "show this" etc, in questions. Good rule of thumb: a question contains a question mark. – Hrodelbert May 18 '15 at 11:24
  • The question has been edited. Please answer it. – anand konjengbam May 18 '15 at 11:32
  • but $2\times2\times2\times1\times1\times1$ is at least $8$... – danimal May 18 '15 at 11:40
  • You are considering only a particular case wherein, the product is $8$. There are cases $($like mentioned above$)$ where the question is contradicted. – anand konjengbam May 18 '15 at 11:42
  • Your counterexample is incorrect: it is true that the $6$ numbers you selected do not have a product of at least $8$, but it is possible to select $6$ numbers which do: the $4$ that are already $32$ together with $2$ of the $6$ whose product is $1$ (since in particular at least $2$ of these must be $\geq1$). Another question: are you sure you want to consider real numbers and not just positive integers (i.e. natural numbers)? – Hrodelbert May 18 '15 at 12:03
  • Yes we are to consider real numbers – anand konjengbam May 18 '15 at 17:18

2 Answers2

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The proof of this is quite easy. You know that $\Pi_{i=0}^9a_i =32$. WLOG assume $a_0\leq a_1\leq \cdots\leq a_9$ (if not swap indexes appropriately). Call $\Pi_{i=4}^9a_i=p$. Then $p\geq 8$. Assume not. Then $a_4<\sqrt[6]{8}$ thus $\forall i\leq 3 (a_i<\sqrt[6]{8})$. But then $q=\Pi_{i=0}^3a_i<(\sqrt[6]{8})^4$ thus $qp<(\sqrt[6]{8})^4*8=4*8=32$. QED

DRF
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  • The "wlog" and "qed" and proof by contradiction likely unsuited to the OP's understanding of logic. The square roots seem unnecessary. We can let $b_i = a_{2i}a_{2i+1}$ for $i=0,\ldots,4$. Then $b_0\leq \cdots \leq b_4,$ and $\prod_{i=0}^4 b_i=\prod_{i=0}^9 a_i=32.$ We rule out $b_4 < 2,$ since otherwise $32 = \prod_{i=0}^4 b_i < \prod_{i=0}^4 2 = 2^5 = 32.$ We rule out $b_0b_1 > 4,$ as otherwise $32 = \prod_{i=0}^4 b_i = (b_0b_1)(b_2b_3)b_4 > 4\cdot 4\cdot 2 = 32.$ Finally deduce from $b_0b_1 \leq 4$ that $\prod_{i=4}^9 a_i = b_2b_3b_4 = 32/(b_0b_1) \geq 32/4 = 8.$ – Tommy R. Jensen Aug 26 '19 at 13:00
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The product of all the numbers in $S$ is $2^5$. If we cube all the numbers in $S$ and multiply them we get $2^{15}$. If we take out the cubes of the four smallest and replace them with the squares of the six largest, we will increase the product. This new product has five copies of each of the largest numbers and is greater than $2^{15}$, so the product of one copy of each of the six largest numbers is greater than $2^3=8$

Ross Millikan
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