If $a + b + c = 3$ and $a > 0, b > 0, c > 0$ , then the greatest value $a^{2} b^{3} c^{2}$ is

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\frac{3^{8} {.2}^{4}}{7^{7}}

\frac{3^{9} {.2}^{4}}{7^{7}}

\frac{3^{10} {.2}^{4}}{7^{7}}

\frac{3^{10} {.2}^{2}}{7^{7}}

answer is A.

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Detailed Solution

\frac{\frac{a}{2} + \frac{a}{2} + \frac{b}{3} + \frac{b}{3} + \frac{b}{3} + \frac{c}{2} + \frac{c}{2}}{7} \geq \sqrt[7]{\frac{a^{2} b^{3} c^{2}}{2^{4} {.3}^{3}}} \Rightarrow \frac{3}{7} \geq \sqrt[7]{\frac{a^{2} b^{3} c^{2}}{2^{4} {.3}^{3}}}

$\Rightarrow {(\frac{3}{7})}^{7} \geq \frac{a^{2} b^{3} c^{2}}{2^{4} {.3}^{3}} \Rightarrow \frac{3^{10} {.2}^{4}}{7^{7}} \geq a^{2} b^{3} c^{2} \Rightarrow a^{2} b^{3} c^{2} \leq \frac{3^{10} {.2}^{4}}{7^{7}}$

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