If a + 2b + 3c = 6,then the greatest value of | vedclass

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(A) Given $a + 2b + 3c = 6$. We want to maximize $abc^2$. We can write this as $a + 2b + \frac{3c}{2} + \frac{3c}{2} = 6$. By the $AM \geq GM$ inequal

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$\frac{9}{8}$

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(A) Given $a + 2b + 3c = 6$. We want to maximize $abc^2$. We can write this as $a + 2b + \frac{3c}{2} + \frac{3c}{2} = 6$. By the $AM \geq GM$ inequality for four positive terms: $\frac{a + 2b + \frac{3c}{2} + \frac{3c}{2}}{4} \geq \sqrt[4]{a \cdot 2b \cdot \frac{3c}{2} \cdot \frac{3c}{2}}$ $\frac{6}{4} \geq \sqrt[4]{a \cdot 2b \cdot \frac{9c^2}{4}}$ $\frac{3}{2} \geq \sqrt[4]{\frac{18}{4} abc^2}$ $\frac{3}{2} \geq \sqrt[4]{\frac{9}{2} abc^2}$ Raising both sides to the power of $4$: $(\frac{3}{2})^4 \geq \frac{9}{2} abc^2$ $\frac{81}{16} \geq \frac{9}{2} abc^2$ $abc^2 \leq \frac{81}{16} \cdot \frac{2}{9} = \frac{9}{8}$.

If $a + 2b + 3c = 6$,then the greatest value of $abc^2$ is (where $a, b, c$ are positive real numbers).

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