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

& 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{aligned}$$ Now, we know that A, B, C are representing angles of triangles ABC and $${{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{aligned} & \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{aligned}$$ Now, we know that value of $$\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{aligned} & \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{aligned}$$ Hence, we can get least value of $$\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.