Let $n\geq 2$ be an integer. Find the smallest real value $\rho (n)$ such that for any $x_i>0$, $i=1,2,\ldots,n$ with $x_1 x_2 \cdots x_n = 1$, the inequality
\[ \sum_{i=1}^n \frac 1{x_i} \leq \sum_{i=1}^n x_i^r \] is true for all $r\geq \rho (n)$.
1. We start with the given inequality:
   \[
   \sum_{i=1}^n \frac{1}{x_i} \leq \sum_{i=1}^n x_i^r
   \]
   where \( x_i > 0 \) for \( i = 1, 2, \ldots, n \) and \( x_1 x_2 \cdots x_n = 1 \).

2. By the Arithmetic Mean-Geometric Mean Inequality (AM-GM Inequality), we have:
   \[
   \frac{x_1 + x_2 + \cdots + x_n}{n} \geq \sqrt[n]{x_1 x_2 \cdots x_n}
   \]
   Given \( x_1 x_2 \cdots x_n = 1 \), it simplifies to:
   \[
   \frac{x_1 + x_2 + \cdots + x_n}{n} \geq 1
   \]
   Therefore:
   \[
   x_1 + x_2 + \cdots + x_n \geq n
   \]

3. Next, we consider the Harmonic Mean-Geometric Mean Inequality (HM-GM Inequality):
   \[
   \frac{n}{\frac{1}{x_1} + \frac{1}{x_2} + \cdots + \frac{1}{x_n}} \leq \sqrt[n]{x_1 x_2 \cdots x_n}
   \]
   Given \( x_1 x_2 \cdots x_n = 1 \), it simplifies to:
   \[
   \frac{n}{\sum_{i=1}^n \frac{1}{x_i}} \leq 1
   \]
   Therefore:
   \[
   \sum_{i=1}^n \frac{1}{x_i} \geq n
   \]

4. To find the smallest \( \rho(n) \), we need to ensure that:
   \[
   \sum_{i=1}^n \frac{1}{x_i} \leq \sum_{i=1}^n x_i^r
   \]
   holds for all \( r \geq \rho(n) \).

5. Consider the case when \( x_1 = x_2 = \cdots = x_{n-1} = \epsilon \) and \( x_n = \frac{1}{\epsilon^{n-1}} \). Then:
   \[
   x_1 x_2 \cdots x_n = \epsilon^{n-1} \cdot \frac{1}{\epsilon^{n-1}} = 1
   \]
   and:
   \[
   \sum_{i=1}^n \frac{1}{x_i} = (n-1) \frac{1}{\epsilon} + \epsilon^{n-1}
   \]
   \[
   \sum_{i=1}^n x_i^r = (n-1) \epsilon^r + \left( \frac{1}{\epsilon^{n-1}} \right)^r = (n-1) \epsilon^r + \frac{1}{\epsilon^{r(n-1)}}
   \]

6. For the inequality to hold, we need:
   \[
   (n-1) \frac{1}{\epsilon} + \epsilon^{n-1} \leq (n-1) \epsilon^r + \frac{1}{\epsilon^{r(n-1)}}
   \]

7. As \( \epsilon \to 0 \), the dominant terms are \( \frac{1}{\epsilon} \) on the left-hand side and \( \frac{1}{\epsilon^{r(n-1)}} \) on the right-hand side. For the inequality to hold, we need:
   \[
   \frac{1}{\epsilon} \leq \frac{1}{\epsilon^{r(n-1)}}
   \]
   which implies:
   \[
   1 \leq \epsilon^{r(n-1) - 1}
   \]
   Since \( \epsilon \to 0 \), this inequality holds if and only if:
   \[
   r(n-1) - 1 \geq 0 \implies r \geq \frac{1}{n-1}
   \]

8. Therefore, the smallest value of \( \rho(n) \) is:
   \[
   \rho(n) = \frac{1}{n-1}
   \]

The final answer is \(\boxed{\frac{1}{n-1}}\).