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 $y_1y_2\cdots y_n\geq 1$?
 A: 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.
A: 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 $G$).
A: A graphical answer:
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.

A: 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 $y_1y_2\cdots y_n \le 1$. 

