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# intermediate algebra

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Let x, y  and z be positive real numbers such that x + y + z = 1. Find the maximum value of xyz.

Jan 28, 2021

#1
+8456
+2

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 = $$\dfrac1{27}$$

Jan 28, 2021

#1
+8456
+2

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 = $$\dfrac1{27}$$

MaxWong Jan 28, 2021