Question

If x, y, z are positive real numbers such that $x+y+z=3$, then find the minimum value of $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}$.

Answer 1

Consider the terms to be $x,y,z$

A.M$=\frac{x+y+z}{3}=1$

G.M $=\sqrt[3]{3xyz}$

$1\ge \sqrt[3]{3xyz}.....\left(1\right)$

Now, consider the terms to be $\frac{1}{x},\frac{1}{y},\frac{1}{z}$

A.M$=\frac{\frac{1}{x}+\frac{1}{y}+\frac{1}{z}}{3}$

G.M $=\sqrt[3]{3\frac{1}{xyz}}...\left(2\right)$

$\left(1\right)\times \left(2\right)$

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

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

Minimum value $=3$