Question

The correct relation between AM, GM and HM is
     A. \[\text{AM}\,\text{=}\,\text{GM}\,\text{=}\,\text{HM}\]
     B. \[\text{AM}\,\le \,\text{GM}\,\le \,\text{HM}\]
     C. \[\text{AM}\,\ge \,\text{GM}\,\ge \,\text{HM}\]
     D. \[\text{AM}\,\text{}\,\text{GM}\,>\,\text{HM}\]

Answer 1

Hint: We will first find the inequality for two numbers a and b, and then extend it to any n numbers using Cauchy induction.

Complete step by step solution: First let’s try to prove that \[\text{AM}\,\ge \,\text{GM}\] for any two numbers a and b, both > 0
We know
     \[\text{GM}\,\text{=}\,\sqrt{\text{ab}}\,\text{=}\,\dfrac{\sqrt{\text{a}}\sqrt{\text{b}}\text{+}\sqrt{\text{b}}\sqrt{\text{a}}}{\text{2}}\,\le \,\dfrac{\sqrt{\text{a}}\sqrt{\text{a}}\text{+}\sqrt{\text{b}}\sqrt{\text{b}}}{\text{2}}\,\text{=}\,\dfrac{\text{a+b}}{\text{2}}\]
Why is $\sqrt{\text{a}}\sqrt{\text{b}}\text{+}\sqrt{\text{b}}\sqrt{\text{a}}\,\text{}\sqrt{\text{a}}\sqrt{\text{a}}\text{+}\sqrt{\text{b}}\sqrt{\text{b}}\,\text{=}\,\text{a+b ?}$
Consider ${{\left( \sqrt{a}-\sqrt{b} \right)}^{2}}$ which we know has to be $\ge \,0$
Expanding,
$\begin{align}
  & a+b-2\sqrt{ab}\,\ge \,0 \\
 & \therefore \,a+b\,\ge \,2\sqrt{ab}\,=\,\sqrt{a}\sqrt{b}+\sqrt{b}\sqrt{a} \\
\end{align}$ .
Now let’s try to prove that $\text{GM}\,\ge \,\text{HM}$.
\[\text{HM}\,\text{of}\,\text{a}\,\text{and}\,\text{b}\,\text{=}\,\dfrac{\text{2ab}}{\text{a+b}}\]
Let us compute $\text{GM}-\text{HM}$
     \[\begin{align}
  & \text{=}\,\sqrt{\text{ab}}\,\text{=}\,\dfrac{\text{2ab}}{\text{(a+b)}} \\
 & \text{=}\,\dfrac{\text{(a+b)}\sqrt{\text{ab}}\text{-2}\sqrt{\text{ab}}\text{.}\sqrt{\text{ab}}}{\text{(a+b)}} \\
 & \text{=}\,\dfrac{\sqrt{\text{ab}}\text{ }\!\![\!\!\text{ a+b }\!\!]\!\!\text{ -2}\sqrt{\text{ab}}}{\text{a+b}} \\
 & \text{=}\,\dfrac{\sqrt{\text{ab}}{{\left[ \sqrt{\text{a}}\text{-}\sqrt{\text{b}} \right]}^{\text{2}}}}{\text{a+b}}
\end{align}\]
Which has to be positive because a, b >0
$\therefore $ if $\begin{align}
  & \text{GM}-\text{HM}\,\ge \,\text{0} \\
 & \,\,\,\,\,\,\,\,\,\,\,\,\,\text{GM}\,\ge \,\text{HM} \\
\end{align}$
So we get the inequality $\text{AM}\,\ge \,\text{GM}\,\ge \,\text{HM}$ for two positive numbers a and b, with the equality occurring when both a and b are equal
Ideally, the solution ends right here and we can choose $\text{AM}\,\ge \,\text{GM}\,\ge \,\text{HM}$ Option C as the right answer. However, for the sake of completion, we will use Cauchy induction to prove $\text{AM}\,\ge \,\text{GM}$ for any N positive numbers. $\text{AM}\,\ge \,\text{GM}$ will be a direct result of $\text{AM}\,\ge \,\text{GM}$. This will be done as an extra exercise in the note.

Note: Note: We know $\text{AM}\,\ge \,\text{GM}\,\ge \,\text{HM}$ for 2 variables. We next prove that if the inequality holds for n variables, then it holds for 2n variables.
Let it hold true for n variables. Let ${{\text{A}}_{\text{n}\,}}\text{,}{{\text{A}}_{\text{n}}}\text{,}\,{{\text{A}}_{\text{2n}}}$ denote arithmetic means of $\left( a,\,...,\,{{a}_{n}} \right)$ q\[\text{(}{{\text{a}}_{\text{n+1}}}\text{,}.....\text{,}\,{{\text{a}}_{\text{2n}}}\text{)}\,\text{,}\,\text{(}{{\text{a}}_{\text{1}}}\text{,}\,...\text{,}\,{{\text{a}}_{\text{2n}}}\text{)}\] respectively; let ${{\text{G}}_{\text{n}}}\text{,}\,{{\text{G}}_{\text{n}}}\text{,}\,{{\text{G}}_{\text{2n}}}$ denote their geometric means. Then
\[{{\text{G}}_{\text{2n}}}\,=\,\sqrt{{{\text{G}}_{\text{n}}}{{\text{G}}_{\text{n}}}}\,\le \,\dfrac{{{\text{G}}_{\text{n}}}\,\text{+}\,{{\text{G}}_{\text{n}}}}{\text{2}}\,\le \,\dfrac{{{\text{A}}_{\text{n}}}{{\text{A}}_{\text{n}}}}{\text{2}}\,=\,{{\text{A}}_{\text{2n}}}\]
Why? $\begin{align}
  & {{G}_{n}}\,=\,\sqrt[n]{{{a}_{1}}.....{{a}_{n}}} \\
 & {{G}_{n}}\,=\,\sqrt[n]{{{a}_{n+1}}.....{{a}_{2n}}} \\
\end{align}$ But \[\begin{align}
  & {{\text{G}}_{\text{2n}}}\,\text{=}\,\sqrt[\text{2n}]{{{\text{a}}_{\text{1}}}.....{{\text{a}}_{\text{2n}}}} \\
 & \text{=}\,{{\left( \,\sqrt[\text{2n}]{{{\text{a}}_{\text{1}}}.....{{\text{a}}_{\text{2n}}}}\text{.}\,\sqrt[\text{2n}]{{{\text{a}}_{\text{n+1}}}.....{{\text{a}}_{\text{2n}}}} \right)}^{{}^{\text{1}}/{}_{\text{2}}}} \\
 & \text{=}\sqrt{{{\text{G}}_{\text{n}}}{{\text{G}}_{\text{n}}}} \\
\end{align}\]
Now \[\sqrt{{{\text{G}}_{\text{n}}}{{\text{G}}_{\text{n}}}\,\le \,\dfrac{{{\text{G}}_{\text{n}}}\,\text{+}\,{{\text{G}}_{\text{n}}}}{\text{2}}}\]
because $\text{AM}\,\ge \,\text{GM}$ holds for ${{\text{G}}_{\text{n}}}$ and ${{\text{G}}_{\text{n}}}$ and $\dfrac{{{\text{A}}_{\text{n}}}\text{+}{{\text{A}}_{\text{n}}}}{\text{2}}\,\ge \,\dfrac{{{\text{G}}_{\text{n}}}\text{+}{{\text{G}}_{\text{n}}}}{\text{2}}$ because ${{\text{A}}_{\text{n}}}\,\ge \,{{\text{G}}_{\text{n}}}$ and \[\text{A}{{\text{ }\!\!'\!\!\text{ }}_{\text{n}}}\,\ge \,\text{G}{{\text{ }\!\!'\!\!\text{ }}_{\text{n}}}\]
These results show by induction holds for \[\text{n}\,\text{=}\,{{\text{2}}^{\text{k}}}\] for all integers k. For every integer n, there is an N > n such that the theorem holds for N variables.
Now finally, we show that if N > n, and the theorem holds for N variables, then it holds for n variables with
$\text{AM}\,\text{=}\,{{\text{A}}_{\text{n}}}\,\text{and}\,\text{GM}\,\text{=}\,{{\text{G}}_{\text{n}}}$
Set ${{\text{b}}_{\text{k}}}\,\text{=}\,{{\text{a}}_{\text{k}}}$ for $1\,\le \,\text{k}\,\le \,\text{n}$ and let ${{\text{b}}_{\text{k}}}\,\text{=}\,{{\text{a}}_{\text{n}}}$ for $\text{n}\,\text{}\,\text{k}\le \,\text{N}$.
Then.
\[G_{n}^{n/N}.\,A_{n}^{(N-n)/N}\,=\,\prod\limits_{i=1}^{N}{b_{i}^{1/N}}\,\le \,\sum\limits_{i=1}^{n}{{{b}_{i/N}}}\,=\,{{A}_{n}}\]
How?
\[\begin{align}
  & {{G}_{n}}\,=\,\prod\limits_{i=1}^{n}{b_{i}^{1/N}} \\
 & {{G}_{N}}\,=\,\,=\,\prod\limits_{i=1}^{N}{b_{i}^{1/N}}=\,\prod\limits_{i=1}^{n}{b_{i}^{1/n}}.\,=\,\prod\limits_{i=1}^{N}{b_{i}^{1/N}} \\
\end{align}\]
But from $i\,=\,n+1$ to $N\,{{b}_{1\,}}=\,{{A}_{n}}$ N for all i
So\[{{G}_{N}}\,=\,\,=\,\prod\limits_{i=1}^{N}{b_{i}^{1/N}}=\,\prod\limits_{i=1}^{n}{b_{i}^{1/n}}.A_{n}^{\dfrac{N-n}{n}}\]
How is\[\,\prod\limits_{i=1}^{N}{b_{i}^{1/N}}\,\le \,\sum\limits_{i=1}^{N}{\dfrac{{{b}_{i}}}{N}}\,?\,\] sunoke yse if $\text{AM}\,\ge \,\text{GM}$
But\[\begin{align}
  & \sum\limits_{\text{i=1}}^{\text{N}}{\dfrac{{{\text{b}}_{\text{i}}}}{\text{N}}}\,\text{=}\,\sum\limits_{\text{i=1}}^{\text{n}}{\dfrac{{{\text{b}}_{\text{i}}}}{\text{N}}}\text{+}\sum\limits_{\text{i=1}}^{\text{N}}{\dfrac{{{\text{b}}_{\text{i}}}}{\text{N}}} \\
 & {{\text{b}}_{\text{i}\,}}\,\text{=}\,{{\text{A}}_{\text{n}}}\,\text{for}\,\text{all}\,\text{i}\,\text{here} \\
\end{align}\]
     \[\begin{align}
  & \text{=}\,\dfrac{\text{1}}{\text{N}}\,\text{ }\!\!\times\!\!\text{ n}{{\text{A}}_{\text{n}}}\,\text{+}\,\dfrac{\text{(N-n)}{{\text{A}}_{\text{n}}}}{\text{N}} \\
 & \text{=}\,\dfrac{\text{N}}{\text{N}}{{\text{A}}_{\text{n}}}\,\text{=}\,{{\text{A}}_{\text{n}}} \\
\end{align}\]
So we have proved $\text{G}_{\text{n}}^{\text{n/N}}\text{A}_{\text{n}}^{\text{(N-n)/N}}\le \,{{\text{A}}_{\text{n}}}$
Cross multiply, we get
$\text{G}_{\text{n}}^{\text{n/N}}\le \text{A}_{\text{n}}^{\text{(N-n)/N}}$

Now, we know $\dfrac{\sum\limits_{i-1}^{n}{\sqrt{\dfrac{{{x}_{1}}{{x}_{2}}...{{x}_{n}}}{{{x}_{i}}n}}}}{n}\ge \,1$ (result of $\text{AM}\,\ge \,\text{GM}$
Which means$\sqrt[\text{n}]{{{\text{x}}_{\text{1}}}{{\text{x}}_{\text{2}}}...{{\text{x}}_{\text{n}}}}\dfrac{\sum\limits_{\text{i=1}}^{\text{n}}{\dfrac{\text{1}}{{{\text{x}}_{\text{1}}}}}}{\text{n}}$
So $\sqrt[\text{n}]{{{\text{x}}_{\text{1}}}{{\text{x}}_{\text{2}}}...{{\text{x}}_{\text{n}}}}\dfrac{\text{n}}{\sum\limits_{\text{i=1}}^{\text{n}}{\dfrac{\text{1}}{{{\text{x}}_{\text{1}}}}}}$ which means $\text{GM}\,\ge \,\text{HM}$