If x, y and z are positive numbers and x + y + z = 1, then the least value of  $$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}$$​is:

 x + y + z = 1 

 x, y, z  are positive numbers.

We are asked to find the minimum value of $$\frac{1}{x} + \frac{1}{y} + \frac{1}{z}$$​.

To minimize$$\frac{1}{x} + \frac{1}{y} + \frac{1}{z}$$​ under the constraint  x + y + z = 1 , we can apply the **Arithmetic Mean-Harmonic Mean (AM-HM) inequality, which states:

$$\frac{x + y + z}{3} \geq \frac{3}{\frac{1}{x} + \frac{1}{y} + \frac{1}{z}}$$​

1. Apply the AM-HM inequality:

Using x + y + z = 1 , we can substitute this into the AM-HM inequality:

$$\frac{1}{3} \geq \frac{3}{\frac{1}{x} + \frac{1}{y} + \frac{1}{z}}$$​

2. Simplify the inequality:

Now, rearrange to solve for $$\frac{1}{x} + \frac{1}{y} + \frac{1}{z}$$​:

$$\frac{1}{x} + \frac{1}{y} + \frac{1}{z} \geq 9$$​

This tells us that the least value of $$\frac{1}{x} + \frac{1}{y} + \frac{1}{z}$$​is at least 9.

3. Determine when equality holds:

The AM-HM inequality achieves equality when  x = y = z . So, for the minimum value to occur, we need  x = y = z .

4. Find the value of x ,  y , and  z :

Since x + y + z = 1  and  x = y = z , we have:

$$3x = 1 \quad \Rightarrow \quad x = \frac{1}{3}$$​

So,  x = y = z = $$\frac{1}{3}$$​.

5. Substitute  x = y = z = $$\frac{1}{3}$$​  into the expression:

Now, calculate  $$\frac{1}{x} + \frac{1}{y} + \frac{1}{z}$$  when  x = y = z = $$\frac{1}{3}$$​ 

$$\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 3 \times \frac{1}{\frac{1}{3}} = 3 \times 3 = 9$$​