Application of AM-GM Inequality - Practice Questions & MCQ

Application of AM-GM
If $a_1,a_2,a_3,\dots \dots .,a_n$ are n positive variables and k is a constant
If $a_1+a_2+a_3+\dots \dots .+a_n=k$ (constant), then the greatest value of $a_1\cdot a_2\cdot a_3$. $$ $\cdot a_n$ is $(\mathrm{k}{)}^{\mathrm{n}}$ and this is possible when $a_1=a_2=a_3=\dots .=a_n$.
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Proof:

$$\begin{array}{rl}& \text{ as, }\mathrm{AM}\ge \mathrm{GM}\\ & \therefore \frac{{\mathrm{a}}_1+{\mathrm{a}}_2+{\mathrm{a}}_3+\dots \dots +{\mathrm{a}}_{\mathrm{n}}}{\mathrm{n}}\ge {({\mathrm{a}}_1\cdot {\mathrm{a}}_2\cdot {\mathrm{a}}_3\cdot \dots \dots \cdot {\mathrm{a}}_{\mathrm{n}})}^{\frac{1}{\mathrm{n}}}\\ & \Rightarrow \frac{\mathrm{k}}{\mathrm{n}}\ge {({\mathrm{a}}_1\cdot {\mathrm{a}}_2\cdot {\mathrm{a}}_3\cdot \dots \dots \cdot {\mathrm{a}}_{\mathrm{n}})}^{\frac{1}{n}}\\ & \text{ or }({\mathrm{a}}_1\cdot {\mathrm{a}}_2\cdot {\mathrm{a}}_3\cdot \dots \dots \cdot {\mathrm{a}}_{\mathrm{n}})\le {\left(\frac{\mathrm{k}}{\mathrm{n}}\right)}^{\mathrm{n}}\end{array}$$
 

Application of AM-GM
$a_1,a_2,a_3,\dots \dots .,a_n$ are n positive variables and k is a constant
If ${\mathrm{a}}_1\cdot {\mathrm{a}}_2\cdot {\mathrm{a}}_3\cdot \dots \dots \cdot {\mathrm{a}}_{\mathrm{n}}=\mathrm{k}$, where k is constant, then the value of ${\mathrm{a}}_1+{\mathrm{a}}_2+{\mathrm{a}}_3+\dots \dots +{\mathrm{a}}_{\mathrm{n}}$ is minimum when all the terms are equal to each other, i.e. ${\mathrm{a}}_1={\mathrm{a}}_2={\mathrm{a}}_3=\dots \dots ={\mathrm{a}}_{\mathrm{n}}$.
So that the least value of $a_1+a_2+a_3+\dots \dots +a_n$ is $n(k{)}^{1/n}$.

Proof:

To prove this we will be using the fact that A.M. $\ge$ G.M

So,

$\begin{array}{rl}& \frac{a_1+a_2+a_3+\dots \dots +a_n}{n}\ge {(a_1\cdot a_2\cdot a_3\cdot \dots \dots \cdot a_n)}^{1/n}=k^{1/n}\\ \Rightarrow & \frac{a_1+a_2+a_3+\dots \dots +a_n}{n}\ge k^{1/n}\\ \Rightarrow & a_1+a_2+a_3+\dots \dots +a_n\ge n\cdot k^{1/n}\end{array}$
Here, $a_1=a_2=a_3=\dots \dots =a_n$
$\therefore$ least value of $a_1+a_2+a_3+\dots \dots +a_n$ is $n\cdot k^{1/n}$

Application of A.M., G.M., and H.M.
Let $\mathrm{A},\mathrm{G}$ and H are arithmetic, geometric and harmonic means of two positive real numbers a and b.

Then,

$\mathrm{A}=\frac{a+b}{2},\mathrm{G}=\sqrt{a\cdot b}\text{ and }\mathrm{H}=\frac{2ab}{a+b}$

1. $A\ge G\ge H$

$\begin{array}{rl}& \mathrm{A}-\mathrm{G}=\frac{a+b}{2}-\sqrt{ab}=\frac{(\sqrt{a}-\sqrt{b}{)}^2}{2}\ge 0\\ & \Rightarrow \mathrm{A}-\mathrm{G}\ge 0\\ & \Rightarrow \mathrm{A}\ge \mathrm{G}\end{array}$

Note that $\mathrm{A}=\mathrm{G}$ when $\mathrm{a}=\mathrm{b}$
Now,

$\begin{array}{rl}& \mathrm{G}-\mathrm{H}=\sqrt{ab}-\frac{2ab}{a+b}\\ & =\sqrt{ab}\left(\frac{a+b-2\sqrt{ab}}{a+b}\right)\\ & =\frac{\sqrt{ab}}{a+b}(\sqrt{a}-\sqrt{b}{)}^2\ge 0\\ & \Rightarrow \mathrm{G}\ge \mathrm{H}\end{array}$
Again $\mathrm{G}=\mathrm{H}$ when $\mathrm{a}=\mathrm{b}$
From (i) and (ii) we get

$\mathrm{A}\ge \mathrm{G}\ge \mathrm{H}$

Note:
- when $\mathbf{a}=\mathrm{b}$ then only, $\mathrm{A}=\mathrm{G}=\mathrm{H}$ If $a_1,a_2,a_3,\dots .,a_n$ are n positive real numbers, then

$\begin{array}{rl}& \mathrm{A}=\text{ A.M. of }{a}_1,a_2,a_3,\dots .,a_n=\frac{a_1+a_2+a_3+\dots .+a_n}{n}\\ & \mathrm{G}=\text{ G.M. of }{a}_1,a_2,a_3,\dots .,a_n={(a_1\cdot a_2\cdot a_3\dots \dots .a_n)}^{\frac{1}{n}}\end{array}$
$\mathrm{H}=\mathrm{H}.\mathrm{M}.\text{ of }{a}_1,a_2,a_3,\dots \dots ,a_n=\frac{n}{\frac{1}{a_1}+\frac{1}{a_2}+\dots \dots +\frac{1}{a_n}}$
In such case also $A\ge G\ge H$
And $\mathrm{A}=\mathrm{G}=\mathrm{H}$, when $a_1=a_2=a_3=\dots \dots .=a_n$
2. $A,G$ and $H$ of 2 positive real numbers form a geometric progression, i.e. $G^2=AH$. we have,

$\begin{array}{rl}\mathrm{A}\cdot \mathrm{H}& =\frac{a+b}{\mathcal{2}}\times \frac{2ab}{a+b}\\ & =ab=(\sqrt{ab}{)}^2={\mathrm{G}}^2\end{array}$
Hence, ${\mathrm{G}}^2=\mathrm{AH}$