Question

In \[\Delta ABC\], least value of \[\dfrac{{{e}^{A}}}{A}+\dfrac{{{e}^{B}}}{B}+\dfrac{{{e}^{C}}}{C}\] is equal to
A.\[\dfrac{9}{\pi }{{e}^{\dfrac{\pi }{3}}}\]
B.\[\dfrac{\pi }{3}{{e}^{\dfrac{\pi }{3}}}\]
C.\[\dfrac{\pi }{9}{{e}^{\dfrac{\pi }{3}}}\]
D.None of these

Answer 1

Hint:Use the property or relation of Arithmetic mean and Geometric mean which can be given as
\[AM\ge GM\](only for positive numbers).Use the fundamental property of a triangle that the sum of interior angles of a triangle is \[{{180}^{\circ }}\]. So, the value of A + B + C is \[180\left( \pi \right)\].

Complete step by step answer:
We know that the Arithmetic mean of n numbers is greater than or equals the Geometric mean of n numbers. Hence, we get
\[\begin{align}
  & A.M\ge G.M \\
 & \dfrac{{{a}_{1}}+{{a}_{2}}+....+{{a}_{n}}}{n}\ge {{\left( {{a}_{1}}.{{a}_{2}}......{{a}_{n}} \right)}^{\dfrac{1}{n}}}-(i) \\
\end{align}\]
Now, we know that A, B, C are representing angles of triangles ABC and \[{{e}^{A}},{{e}^{B}}\] and \[{{e}^{C}}\] will also be a positive value.
Hence, we can apply \[AM\ge GM\]relation with the values \[\dfrac{{{e}^{A}}}{A},\dfrac{{{e}^{B}}}{B},\dfrac{{{e}^{C}}}{C}\] from the equation (i), So we get,
\[\dfrac{\dfrac{{{e}^{A}}}{A}+\dfrac{{{e}^{B}}}{B}+\dfrac{{{e}^{C}}}{C}}{3}\ge {{\left( \dfrac{{{e}^{A}}}{A}.\dfrac{{{e}^{B}}}{B}.\dfrac{{{e}^{C}}}{C} \right)}^{\dfrac{1}{3}}}\]
\[\Rightarrow \dfrac{{{e}^{A}}}{A}+\dfrac{{{e}^{B}}}{B}+\dfrac{{{e}^{C}}}{C}\ge 3{{\left( \dfrac{{{e}^{A}}}{A}.\dfrac{{{e}^{B}}}{B}.\dfrac{{{e}^{C}}}{C} \right)}^{\dfrac{1}{3}}}-(ii)\]
Now, we can use property of surds given as,
\[{{a}^{m}}.{{a}^{n}}={{a}^{m+n}}\]
Hence, we can re – write the equation (ii) as
\[\dfrac{{{e}^{A}}}{A}+\dfrac{{{e}^{B}}}{B}+\dfrac{{{e}^{C}}}{C}\ge 3{{\left( \dfrac{{{e}^{A+B+C}}}{ABC} \right)}^{\dfrac{1}{3}}}\]
Now we know the fundamental property of a triangle is given as “sum of all interior angles of a triangle is \[{{180}^{\circ }}\]”.
Hence, we can replace A + B + C in the above equation by \[{{180}^{\circ }}\]or \[\pi \](radian), we get
\[\dfrac{{{e}^{A}}}{A}+\dfrac{{{e}^{B}}}{B}+\dfrac{{{e}^{C}}}{C}\ge 3{{\left( \dfrac{{{e}^{\pi }}}{ABC} \right)}^{\dfrac{1}{3}}}\]
\[\Rightarrow \dfrac{{{e}^{A}}}{A}+\dfrac{{{e}^{B}}}{B}+\dfrac{{{e}^{C}}}{C}\ge 3\dfrac{{{\left( {{e}^{\pi }} \right)}^{\dfrac{1}{3}}}}{{{\left( ABC \right)}^{\dfrac{1}{3}}}}\]
Now we can use property of surds \[{{\left( {{a}^{m}} \right)}^{n}}={{a}^{mn}}\] in the above equation with the term \[{{\left( {{e}^{\pi }} \right)}^{\dfrac{1}{3}}}\]; So, we get
\[\dfrac{{{e}^{A}}}{A}+\dfrac{{{e}^{B}}}{B}+\dfrac{{{e}^{C}}}{C}\ge 3\dfrac{{{e}^{\dfrac{\pi }{3}}}}{{{\left( ABC \right)}^{\dfrac{1}{3}}}}-(iii)\]
Now A, B, C are angles of triangles so we can apply property \[AM\ge GM\](equation (i)) with the values A, B, C. Hence, we get
\[\dfrac{A+B+C}{3}\ge {{\left( ABC \right)}^{\dfrac{1}{3}}}-(iv)\]
Now, put \[A+B+C=\pi \]using the fundamental property of a triangle. Hence, we can simplify equation (iv) as
\[\dfrac{\pi }{3}\ge {{\left( ABC \right)}^{\dfrac{1}{3}}}\]
\[\Rightarrow \dfrac{\pi }{3}\ge \dfrac{{{\left( ABC \right)}^{\dfrac{1}{3}}}}{1}\]
Now, we can change the inequality of above equation by reversing the fractions as we know,
If \[\dfrac{a}{b}>\dfrac{c}{d}\] then \[\dfrac{b}{a}>\dfrac{d}{c}\], we get
\[\dfrac{3}{\pi }\le \dfrac{1}{{{\left( ABC \right)}^{\dfrac{1}{3}}}}\].
Now, multiply by \[3{{e}^{\dfrac{\pi }{3}}}\] on both sides of the above equation and inequality will not change because \[3{{e}^{\dfrac{\pi }{3}}}\] is a positive value as angle of \[{{e}^{x}}\]is \[\left( 0,\infty \right)\], we get
\[\begin{align}
  & \dfrac{3}{\pi }\times 3{{e}^{\dfrac{\pi }{3}}}\le \dfrac{3{{e}^{\dfrac{\pi }{3}}}}{{{\left( ABC \right)}^{\dfrac{1}{3}}}} \\
 & \dfrac{9}{\pi }{{e}^{\dfrac{\pi }{3}}}\le \dfrac{3{{e}^{\dfrac{\pi }{3}}}}{{{\left( ABC \right)}^{\dfrac{1}{3}}}}-(v) \\
\end{align}\]
Now, we know that value of \[\dfrac{{{e}^{A}}}{A}+\dfrac{{{e}^{B}}}{B}+\dfrac{{{e}^{C}}}{C}\] is greater than \[\dfrac{3{{e}^{\dfrac{\pi }{3}}}}{{{\left( ABC \right)}^{\dfrac{1}{3}}}}\] from equation (iii) and value of \[\dfrac{3{{e}^{\dfrac{\pi }{3}}}}{{{\left( ABC \right)}^{\dfrac{1}{3}}}}\] is greater than \[\dfrac{9}{\pi }{{e}^{\dfrac{\pi }{3}}}\] from the equation (v). Hence, we can write the equation combining the equations (iii) and (v) as,
\[\begin{align}
  & \dfrac{3}{\pi }\times 3{{e}^{\dfrac{\pi }{3}}}\le \dfrac{3{{e}^{\dfrac{\pi }{3}}}}{{{\left( ABC \right)}^{\dfrac{1}{3}}}} \\
 & \dfrac{{{e}^{A}}}{A}+\dfrac{{{e}^{B}}}{B}+\dfrac{{{e}^{C}}}{C}\ge \dfrac{3{{e}^{\dfrac{\pi }{3}}}}{{{\left( ABC \right)}^{\dfrac{1}{3}}}}\ge \dfrac{9}{\pi }{{e}^{\dfrac{\pi }{3}}} \\
\end{align}\]
Hence, we can get least value of \[\dfrac{{{e}^{A}}}{A}+\dfrac{{{e}^{B}}}{B}+\dfrac{{{e}^{C}}}{C}\] by writing the inequality between first term and last term. Hence, we get
\[\dfrac{{{e}^{A}}}{A}+\dfrac{{{e}^{B}}}{B}+\dfrac{{{e}^{C}}}{C}\ge \dfrac{9}{\pi }{{e}^{\dfrac{\pi }{3}}}\]
So, least value of \[\dfrac{{{e}^{A}}}{A}+\dfrac{{{e}^{B}}}{B}+\dfrac{{{e}^{C}}}{C}\] is \[\dfrac{9}{\pi }{{e}^{\dfrac{\pi }{3}}}\].


Note: One may try to calculate exact value of \[\dfrac{{{e}^{A}}}{A}+\dfrac{{{e}^{B}}}{B}+\dfrac{{{e}^{C}}}{C}\], which will be very complex approach and difficult too. As we need only the least value of the given expression, so we can relate it with \[AM\ge GM\] and it is the key point of the question as well.
Applying \[AM\ge GM\] property twice is the key point of the question.