If x, y and z are positive numbers and x + y + z = 1, then the least value of  1x+1y+1z\frac{1}{x}+\frac{1}{y}+\frac{1}{z}x1+y1+z1is:

Correct option is C

Given:

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}

Formula Used:

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}}

Solution :

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

Ans 9

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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:

7

5

9

11

Correct option is C