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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

Best Answer 

 #1
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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
avatar+8461 
+2
Best Answer

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

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