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Question: The correct relation between AM, GM and HM is A. \[\text{AM}\,\text{=}\,\text{GM}\,\text{=}\,...

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

Explanation

Solution

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 AM ≥ GM\text{AM}\,\ge \,\text{GM} for any two numbers a and b, both > 0
We know
GM = ab = ab+ba2 ≤ aa+bb2 = a+b2\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 ab+ba aa+bb = a+b ?\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 (a−b)2{{\left( \sqrt{a}-\sqrt{b} \right)}^{2}} which we know has to be ≥ 0\ge \,0
Expanding,
a+b−2ab ≥ 0 ∴ a+b ≥ 2ab = ab+ba \begin{aligned} & a+b-2\sqrt{ab}\,\ge \,0 \\\ & \therefore \,a+b\,\ge \,2\sqrt{ab}\,=\,\sqrt{a}\sqrt{b}+\sqrt{b}\sqrt{a} \\\ \end{aligned} .
Now let’s try to prove that GM ≥ HM\text{GM}\,\ge \,\text{HM}.
HM of a and b = 2aba+b\text{HM}\,\text{of}\,\text{a}\,\text{and}\,\text{b}\,\text{=}\,\dfrac{\text{2ab}}{\text{a+b}}
Let us compute GM−HM\text{GM}-\text{HM}
= ab = 2ab(a+b) = (a+b)ab-2ab.ab(a+b) = ab !![!! a+b !!]!! -2aba+b = ab[a-b]2a+b\begin{aligned} & \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{aligned}
Which has to be positive because a, b >0
∴\therefore if GM−HM ≥ 0              GM ≥ HM \begin{aligned} & \text{GM}-\text{HM}\,\ge \,\text{0} \\\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\text{GM}\,\ge \,\text{HM} \\\ \end{aligned}
So we get the inequality AM ≥ GM ≥ HM\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 AM ≥ GM ≥ HM\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 AM ≥ GM\text{AM}\,\ge \,\text{GM} for any N positive numbers. AM ≥ GM\text{AM}\,\ge \,\text{GM} will be a direct result of AM ≥ GM\text{AM}\,\ge \,\text{GM}. This will be done as an extra exercise in the note.

Note: Note: We know AM ≥ GM ≥ HM\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 An ,An, A2n{{\text{A}}_{\text{n}\,}}\text{,}{{\text{A}}_{\text{n}}}\text{,}\,{{\text{A}}_{\text{2n}}} denote arithmetic means of (a, ..., an)\left( a,\,...,\,{{a}_{n}} \right) q(an+1,....., a2n) , (a1, ..., a2n)\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 Gn, Gn, G2n{{\text{G}}_{\text{n}}}\text{,}\,{{\text{G}}_{\text{n}}}\text{,}\,{{\text{G}}_{\text{2n}}} denote their geometric means. Then
G2n = GnGn ≤ Gn + Gn2 ≤ AnAn2 = A2n{{\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? Gn = a1.....ann Gn = an+1.....a2nn \begin{aligned} & {{G}_{n}}\,=\,\sqrt[n]{{{a}_{1}}.....{{a}_{n}}} \\\ & {{G}_{n}}\,=\,\sqrt[n]{{{a}_{n+1}}.....{{a}_{2n}}} \\\ \end{aligned} But G2n = a1.....a2n2n = ( a1.....a2n2n. an+1.....a2n2n)1/2 =GnGn \begin{aligned} & {{\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{aligned}
Now GnGn ≤ Gn + Gn2\sqrt{{{\text{G}}_{\text{n}}}{{\text{G}}_{\text{n}}}\,\le \,\dfrac{{{\text{G}}_{\text{n}}}\,\text{+}\,{{\text{G}}_{\text{n}}}}{\text{2}}}
because AM ≥ GM\text{AM}\,\ge \,\text{GM} holds for Gn{{\text{G}}_{\text{n}}} and Gn{{\text{G}}_{\text{n}}} and An+An2 ≥ Gn+Gn2\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 An ≥ Gn{{\text{A}}_{\text{n}}}\,\ge \,{{\text{G}}_{\text{n}}} and A !!′!! n ≥ G !!′!! n\text{A}{{\text{ }\\!\\!'\\!\\!\text{ }}_{\text{n}}}\,\ge \,\text{G}{{\text{ }\\!\\!'\\!\\!\text{ }}_{\text{n}}}
These results show by induction holds for n = 2k\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
AM = An and GM = Gn\text{AM}\,\text{=}\,{{\text{A}}_{\text{n}}}\,\text{and}\,\text{GM}\,\text{=}\,{{\text{G}}_{\text{n}}}
Set bk = ak{{\text{b}}_{\text{k}}}\,\text{=}\,{{\text{a}}_{\text{k}}} for 1 ≤ k ≤ n1\,\le \,\text{k}\,\le \,\text{n} and let bk = an{{\text{b}}_{\text{k}}}\,\text{=}\,{{\text{a}}_{\text{n}}} for n  k≤ N\text{n}\,\text{}\,\text{k}\le \,\text{N}.
Then.
Gnn/N. An(N−n)/N = ∏i=1Nbi1/N ≤ ∑i=1nbi/N = AnG_{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?

& {{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{aligned}$$ But from $i\,=\,n+1$ to $N\,{{b}_{1\,}}=\,{{A}_{n}}$ N for all i So$${{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{aligned} & \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{aligned}$$ $$\begin{aligned} & \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{aligned}$$ 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 $\text{GM}\,\ge \,\text{HM}$