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