A) \[\frac{\pi }{3}\,{{e}^{\pi /9}}\]
B) \[\frac{\pi }{9}\,{{e}^{\pi /3}}\]
C) \[\frac{9}{\pi }\,{{e}^{\pi /3}}\]
D) \[\frac{3}{\pi }\,{{e}^{\pi /9}}\]
Correct Answer: C
Solution :
We have, \[\because \] \[AM\ge GM\] \[\frac{{{e}^{A}}}{A}+\frac{{{e}^{B}}}{B}+\frac{{{e}^{C}}}{C}\ge 3\,{{\left( \frac{{{e}^{A+B+C}}}{ABC} \right)}^{1/3}}\] \[=3\,{{\left( \frac{{{e}^{\pi }}}{ABC} \right)}^{1/3}}\] ?(i) and \[A+B+C\ge 3\,\,{{(ABC)}^{1/3}}\] \[\Rightarrow \] \[\frac{\pi }{3}\ge \,{{(ABC)}^{1/3}}\] \[\Rightarrow \] \[\frac{\pi }{3}\le \frac{1}{{{(ABC)}^{1/3}}}\] \[\Rightarrow \] \[\frac{3}{\pi }\,{{({{e}^{\pi }})}^{1/3}}\le {{\left( \frac{{{e}^{\pi }}}{ABC} \right)}^{1/3}}\] \[\Rightarrow \] \[3\left( \frac{3}{\pi } \right)\,{{e}^{\pi }}^{/3}\le 3{{\left( \frac{{{e}^{\pi }}}{ABC} \right)}^{1/3}}\] \[\Rightarrow \] \[\frac{{{e}^{A}}}{A}+\frac{{{e}^{B}}}{B}+\frac{{{e}^{C}}}{C}\ge 3\cdot \,\left( \frac{3}{\pi } \right){{e}^{\pi /3}}=\frac{9}{\pi }\,{{e}^{\pi /3}}\] [from Eq. (i)] \[\Rightarrow \] Least vlaue is \[\frac{9}{\pi }\,{{e}^{\pi /3}}\].
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