Question

If $x+y+z=1$, then the least value of $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}$, is

• A. $3$
• B. $9$
• C. $1$
• D. $27$

Answer 1

Answer: (B)

$9$

Given, $x+y+z=1$

We know that $A.M.\ge G.M.$ and $G.M.\ge H.M.$

therefore $A.M.\ge H.M.$

$⟹\frac{x+y+z}{3}\ge \frac{3}{\frac{1}{x}+\frac{1}{y}+\frac{1}{z}}$

$⟹\frac{1}{3}\ge \frac{3}{\frac{1}{x}+\frac{1}{y}+\frac{1}{z}}$

$⟹\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\ge 9$

Therefore the least value of $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}$ is $9$