Question

If the arithmetic, geometric and harmonic means between two distinct positive real numbers be $A$, $G$ and $H$ respectively, then the relation between them is
(1) $A > G > H$
(2) $A > G < H$
(3) $H > G > A$
(4) $G > A > H$

Answer 1

In this type of question we have to use formulas of different types of means. We know that the three Pythagorean means are the arithmetic mean (AM), geometric mean (GM) and the harmonic mean (HM). Also we know that if $a$ and $b$ are two positive numbers then Arithmetic Mean (AM) = $\dfrac{\left( a+b \right)}{2}$, Geometric Mean (GM) = $\sqrt{ab}$ and Harmonic Mean (HM) = $\dfrac{2ab}{\left( a+b \right)}$.

Complete step-by-step solution:
Now we have to find the relation between the arithmetic, geometric and harmonic means between two distinct positive real numbers which are represented by $A$, $G$ and $H$ respectively.
Let us assume $a$ and $b$ be the two distinct positive real numbers then we have

& \Rightarrow AM=A=\dfrac{\left( a+b \right)}{2} \\\ & \Rightarrow GM=G=\sqrt{ab} \\\ & \Rightarrow HM=H=\dfrac{2ab}{\left( a+b \right)} \\\ \end{aligned}$$ Now to find the relation we have to consider following expression $$\begin{aligned} & \Rightarrow A-G=\dfrac{\left( a+b \right)}{2}-\sqrt{ab} \\\ & \Rightarrow A-G=\dfrac{\left( a+b \right)-2\sqrt{ab}}{2} \\\ & \Rightarrow A-G=\dfrac{{{\left( \sqrt{a} \right)}^{2}}+{{\left( \sqrt{b} \right)}^{2}}-2\sqrt{a}\sqrt{b}}{2} \\\ & \Rightarrow A-G=\dfrac{{{\left( \sqrt{a}-\sqrt{b} \right)}^{2}}}{2} \\\ & \Rightarrow A-G\ge 0 \\\ & \Rightarrow A\ge G\cdots \cdots \cdots \left( i \right) \\\ \end{aligned}$$ Also let us consider the another expression $$\begin{aligned} & \Rightarrow G-H=\sqrt{ab}-\dfrac{2ab}{\left( a+b \right)} \\\ & \Rightarrow G-H=\dfrac{\sqrt{ab}\left( a+b \right)-2ab}{\left( a+b \right)} \\\ & \Rightarrow G-H=\dfrac{\left( \sqrt{ab} \right)\left[ \left( a+b \right)-2\sqrt{ab} \right]}{\left( a+b \right)} \\\ & \Rightarrow G-H=\dfrac{\left( \sqrt{ab} \right)\left[ {{\left( \sqrt{a} \right)}^{2}}+{{\left( \sqrt{b} \right)}^{2}}-2\sqrt{a}\sqrt{b} \right]}{\left( a+b \right)} \\\ & \Rightarrow G-H=\dfrac{\left( \sqrt{ab} \right){{\left( \sqrt{a}-\sqrt{b} \right)}^{2}}}{\left( a+b \right)} \\\ & \Rightarrow G-H\ge 0 \\\ & \Rightarrow G\ge H\cdots \cdots \cdots \left( ii \right) \\\ \end{aligned}$$ Hence from equation $$\left( i \right)$$ and $$\left( ii \right)$$ we get $$\Rightarrow A\ge G\ge H$$ **Thus, option (1) is the correct option.** **Note:** In this type of question students have to remember the formulas of arithmetic mean, geometric mean and harmonic mean. Students have to take care during the simplification of the difference between AM and GM as well as between GM and HM. Also students have to note that as both the differences are in the form of squares obviously the differences are greater than or equal to zero.