Question

If x, y, z are positive, and x + y + z = 1, then the maximum value of xyz is

• A. $\frac{1}{3}$
• B. $\frac{1}{5}$
• C. $\frac{1}{25}$
• D. $\frac{1}{27}$

Answer 1

The correct option is D

$\frac{1}{27}$

For the 3 variables x,y,z

AM $\ge$ GM

$\Rightarrow$ $\frac{x+y+z}{3}$ $\ge$ $(xyz{)}^{\frac{1}{3}}$

$\Rightarrow$ $\frac{1}{3}$ $\ge$ $(xyz{)}^{\frac{1}{3}}$

$\Rightarrow$ $(xyz{)}^{\frac{1}{3}}$ $\le$ $\frac{1}{3}$

$\Rightarrow$ xyz $\le$ $\frac{1}{27}$

So, the maximum value of xyz is $\frac{1}{27}$