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# HELP.

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Find the minimum value of the product P(x,y,z)=(2x+3y)(x+3z)(y+2z), when xyz=1 and x,y,z are positive real numbers.

Jun 23, 2020

#1
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Find the minimum value of the product $$P(x,y,z)=(2x+3y)(x+3z)(y+2z)$$, when $$xyz=1$$ and $$x$$, $$y$$, $$z$$ are positive real numbers.

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

$$\begin{array}{|rcll|} \hline \dfrac{2x+3y}{2} & \ge & \sqrt{2x3y} \\\\ \dfrac{x+3z}{2} & \ge & \sqrt{x3z} \\\\ \dfrac{y+2z}{2} & \ge & \sqrt{y2z} \\\\ \hline \left( \dfrac{2x+3y}{2} \right) \left( \dfrac{x+3z}{2} \right) \left( \dfrac{y+2z}{2} \right) & \ge & \sqrt{6xy}\sqrt{3xz} \sqrt{2yz} \\\\ \dfrac{(2x+3y)(x+3z)(y+2z)}{8} & \ge & \sqrt{36x^2y^2z^2} \\\\ \dfrac{(2x+3y)(x+3z)(y+2z)}{8} & \ge & 6xyz \quad | \quad xyz=1 \\\\ \dfrac{(2x+3y)(x+3z)(y+2z)}{8} & \ge & 6*1 \\\\ \dfrac{(2x+3y)(x+3z)(y+2z)}{8} & \ge & 6 \\\\ (2x+3y)(x+3z)(y+2z) & \ge & 6 *8 \\\\ (2x+3y)(x+3z)(y+2z) & \ge & 48 \\\\ \mathbf{ 48 } & \le & \mathbf{(2x+3y)(x+3z)(y+2z)} \\ \hline \end{array}$$

The minimum value is $$\mathbf{ 48 }$$

Jun 23, 2020