If $x, y, z > 0$ and $x + 2 y + 3 z = 2,$ then find the maximum value of $x y z^{2}$

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Solution

Given $x + 2 y + 3 z = 2$
$⟹ x + 2 y + \frac{3 z}{2} + \frac{3 z}{2} = 2$
$A . M \geq G . M$
$⟹ \frac{x + 2 y + \frac{3 z}{2} + \frac{3 z}{2}}{4} \geq \sqrt[4]{(x) (2 y) (3 z / 2)^{2}}$
$⟹ \frac{2}{4} \geq \sqrt[4]{\frac{9}{2} (x y z^{2})}$
$\frac{9}{2} x y z^{2} \leq \frac{1}{16}$
$x y z^{2} \leq \frac{1}{72}$
The maximum value of $x y z^{2}$ is $\frac{1}{72}$

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