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# help

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If x, y, z are positive real numbers such that xyz = 8, find the minimum value of x + y + z.

Dec 4, 2019

#1
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If x, y, z are positive real numbers such that xyz = 8, find the minimum value of x + y + z.

$$\mathbf{\huge{AM \geq GM }}$$

$$\begin{array}{|rcll|} \hline \dfrac{x+y+z}{3} &\geq& \sqrt[3]{xyz} = \sqrt[3]{8} \\ \dfrac{x+y+z}{3} &\geq& \sqrt[3]{8} \\ \dfrac{x+y+z}{3} &\geq& \sqrt[3]{2^3} \\ \dfrac{x+y+z}{3} &\geq& 2 \quad | \quad *3 \\ \mathbf{ x+y+z } &\geq& \mathbf{6} \\ \hline \end{array}$$

The minimum value of $$x + y + z$$ is 6

Thank you, Guest !

Dec 4, 2019
edited by heureka  Dec 4, 2019
#2
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AM.

Do we not have to divide by 3 ?

Guest Dec 4, 2019