# If $\sum\limits_{k=1}^n y_k\geq n$ and $\sum\limits_{k=1}^n \frac{1}{y_k}\geq n$, then $\prod\limits_{k=1}^n y_k\geq 1$?

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 $y_1y_2\cdots y_n\geq 1$?

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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 $y_1y_2 =\frac 34<1$. –  Davide Giraudo Jun 22 '11 at 13:22
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 $y_n < 1/(y_1\dots y_{n-1})$. –  ShreevatsaR Jun 22 '11 at 13:24
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.) –  amWhy Jun 22 '11 at 13:39
@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. –  Dactyl Jun 22 '11 at 15:01
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.) –  amWhy Jun 22 '11 at 16:19

Let $a>0$ be any number.

If $n \geq 3$, just pick $y_1=n$ and $y_2=\frac{1}{n}$, $y_3=a$ and $y_4=..=y_n=1$.

Then $y_1+\ldots+y_n\geq n$ and $1/y_1+\ldots+1/y_n\geq n$.

Anyhow $y_1y_2...y_n =a$. Thus the product can be anything.

If $n=1$, then the problem is trivial.

If $n=2$, then some counterexamples were already provided. Anyhow, if $a<1$ pick $y_1=2$ and $y_2=\frac{a}{2}$; while if $a>1$, pick $y_1=2a$ and $y_2=\frac{1}{2}$, and again the product is $a$. I let out the trivial case $a=1$.

So give those conditions, as long as $n \geq 2$, the product can be any positive number.

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Let $A = A(y_1,y_2 \cdots y_n)$ be the arithmetic mean of the positive numbers, and analogously $G$ and $H$ the geometric and harmonic means.

Your conditions are equivalent to $A\ge 1$, $H \le 1$. And you are asking if we can conclude from that that $G \ge 1$ .

But what we know is that (in general) the geometric mean lies between the others: $H \le G \le A$ . So, we cannot conclude that (or anything about $G$).

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Note that if the conclusion is not true for $n=2$, we can reduce it to this case taking all the other $n-2$ numbers to be $1$.

Below is a graph of the functions $x+y=2$ (red) $\frac{1}{x}+\frac{1}{y}=2$ (black) and $xy=1$ (blue). Our conditions imply that the searched points are above the red graph and below the black one (the region between $x+y=2$ and $\frac{1}{x}+\frac{1}{y}=2$. Note that the region contains points both above and below the graph of $xy=1$, which tells us that we cannot conclude that the product $xy$ is greater than 1.

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It is not true, as girdav's and ShreevatsaR's counter-examples show.

What is true is that the geometric mean is between the arithmetic mean and the harmonic mean of positve numbers, i.e.

$$\frac{n}{1/y_1+1/y_2\ldots+1/y_n} \le \sqrt[n]{y_1y_2\cdots y_n} \le \frac{y_1+y_2+\ldots+y_n}{n}$$

In particular,

• if $1/y_1+1/y_2\ldots+1/y_n \le n$ then $y_1y_2\cdots y_n \ge 1$,
• if $y_1+y_2+\ldots+y_n \le n$ then $y_1y_2\cdots y_n \le 1$.
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