If $x + y + z = 6, (x, y, z > 0)$ , find the maximum value of xyz.

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Solution

given $x + y + z = 6 (x, y, z > 0)$
We know that for a finite number of terms (real)
A.M $\geq$ G.M
Consider the terms to be $x, y, z$
A.M $= \frac{x + y + z}{3} = 2$
G.M $= \sqrt[3]{x y z}$
A.M $\geq$ G.M
$2 \geq \sqrt[3]{x y z}$
$x y z \leq 8$
The maximum value is $8$

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