Question

The least value of $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}$ for positive values of $x,y,z$ such that $x+y+z=9$ is

• A. $0$
• B. $1$
• C. $2$
• D. $4$

Answer 1

The correct option is B

$1$

Explanation of the correct option:

Given: $x+y+z=9$ and $x,y,z$ are positive values.

We know that Arithmetic Mean$\ge$Harmonic Mean

Thus, $\frac{x+y+z}{3}\ge \frac{3}{\left({\displaystyle \frac{1}{x}}+{\displaystyle \frac{1}{y}}+{\displaystyle \frac{1}{z}}\right)}$

$\Rightarrow$ $\frac{9}{3}\ge \frac{3}{\left({\displaystyle \frac{1}{x}}+{\displaystyle \frac{1}{y}}+{\displaystyle \frac{1}{z}}\right)}$

$\Rightarrow$ $1\ge \frac{1}{\left({\displaystyle \frac{1}{x}}+{\displaystyle \frac{1}{y}}+{\displaystyle \frac{1}{z}}\right)}$

$\Rightarrow$$\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\ge 1$

Therefore, the least value of $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}$ is $1$.

Hence option B is the correct option.