If `x_i>0` i=1,2,3,...50 `x_1+x_2...+x_50=50` then the minimum value of `1/(x_1) +1/(x_2) +...+1/(x_50)` equals

Options

A. 50
B. `(50)^(2)`
C. `(50)^(3)`
D. `(50)^(4)`

Answer

To find the minimum value of \( \frac{1}{x_1} + \frac{1}{x_2} + \ldots + \frac{1}{x_{50}} \) given that \( x_i > 0 \) for \( i = 1, 2, \ldots, 50 \) and \( x_1 + x_2 + \ldots + x_{50} = 50 \), we can use the concept of the Arithmetic Mean-Harmonic Mean (AM-HM) inequality.

### Step-by-step Solution:

1. **Understanding the Problem**:
We need to minimize the sum of the reciprocals of \( x_i \) values while the sum of \( x_i \) values equals 50.

2. **Apply the AM-HM Inequality**:
The AM-HM inequality states that for any set of positive numbers \( a_1, a_2, \ldots, a_n \):
\[ \frac{a_1 + a_2 + \ldots + a_n}{n} \geq \frac{n}{\frac{1}{a_1} + \frac{1}{a_2} + \ldots + \frac{1}{a_n}} \]
In our case, let \( a_i = x_i \) for \( i = 1, 2, \ldots, 50 \).

3. **Calculate the Arithmetic Mean**:
We know:
\[ \frac{x_1 + x_2 + \ldots + x_{50}}{50} = \frac{50}{50} = 1 \]
Therefore, by AM-HM:
\[ 1 \geq \frac{50}{\frac{1}{x_1} + \frac{1}{x_2} + \ldots + \frac{1}{x_{50}}} \]

4. **Rearranging the Inequality**:
Rearranging gives us:
\[ \frac{1}{x_1} + \frac{1}{x_2} + \ldots + \frac{1}{x_{50}} \geq 50 \]

5. **Conclusion**:
The minimum value of \( \frac{1}{x_1} + \frac{1}{x_2} + \ldots + \frac{1}{x_{50}} \) is therefore 50.

### Final Answer:
The minimum value of \( \frac{1}{x_1} + \frac{1}{x_2} + \ldots + \frac{1}{x_{50}} \) is **50**.