If $x + y + z = 6 (x, y, z > 0)$ . Find the maximum value of $x^{2} y z$ .

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Solution

$x + y + z = 6 (\frac{x}{2}) + (\frac{x}{2}) + y + z = 6 u s e A M \geq G M (\frac{(\frac{x}{2}) + (\frac{x}{2}) + y + z}{4}) \geq \sqrt{(\frac{x}{2}) (\frac{x}{2}) y z} (\frac{6}{4}) \geq \sqrt{(\frac{x^{2} y z}{4})} ∴ (\frac{x^{2} y z}{4}) \leq (\frac{36}{16}) x^{2} y z \leq 9$
Maximum value is 9.

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