3
$\begingroup$

Let $y_1,\ldots y_n$ be positive real numbers satisfying

$y_1+\cdots+y_n\geq n$ and $\displaystyle{\frac{1}{y_1}+\cdots+\frac{1}{y_n}\geq n}$.

Is it true that $y_1y_2\cdots y_n\geq 1$?

  • 1
    For $n=2$, $y_1=\frac 12$ and $y_2=\frac 32$ we have $y_1+y_2 =2$ and $\frac 1{y_1}+\frac 1{y_2} = 2+\frac 23\geq 2$ but $y_1y_2 =\frac 34<1$.2011-06-22
  • 1
    No, it's not true. For instance, $n=2$, $y_1 = 2$, $y_2 = 1/3$. More generally, you can always take any set of values such that $y_1 + \dots + y_{n-1} \ge n$ and $1/y_1 + \dots + 1/y_{n-1} \ge n$, and then take $y_n < 1/(y_1\dots y_{n-1})$.2011-06-22
  • 0
    Just curious: how did your conjecture originate? Sincerely, what led you to speculate that the inequality might be true? (no sarcasm intended, I'm really curious: was this a homework problem, (e.g., an exercise in proof methods (e.g. refutation by counter-example) did you see some sort of a problem?, etc.)2011-06-22
  • 0
    @amWhy: I think the heart of the AM-GM-HM inequality is the fact that for any set of positive numbers, if their product is 1, then their sum is greater than the order of the set (the AM-GM-HM inequality is a trivial consequence of this). Looked at this way, this question is very natural.2011-06-22
  • 0
    Thanks, @Dactyl: I suspect you're right. I certainly didn't mean to imply that the question came out of nowhere, so to speak...It's more about trying to understand misunderstandings (I tutor, and I'm most helpful when I understand the thought processes that generate questions, and at times, errors.)2011-06-22

4 Answers 4