Let x, y and z be positive real numbers such that x + y + z = 1. Find the maximum value of xyz.
x + y + z = 1.
\(\dfrac{x + y + z}3 = \dfrac13\).
By AM-GM inequality,
\(\dfrac{x + y +z}3 \geq \sqrt[3]{xyz}\\ \sqrt[3]{xyz} \leq\dfrac13\\ xyz \leq \dfrac1{27}\)
Maximum value = \(\dfrac1{27}\)
