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# Relationship between Arithematic and Geometric Means

**Relation:** For A being the arithematic mean and G being the Geometric mean of positive real numbers **a** and **b** then we habe A= $\frac{a+b}{2}$ and G= $\sqrt {ab}$ then
$$A \geq G$$

**Proof:**
Given two positive numbers **a** and **b** and **A** and **G** are arithematic and geometric means respectively.

$$A-G=\frac{a+b}{2}-\sqrt {ab}$$

$$A-G=\frac{a+b-2\sqrt {ab}}{2}$$

$$A-G=\frac{(\sqrt a)^2+(\sqrt b)^2-2\sqrt {ab}}{2}$$

using the algebraic formula $a^2+b^2-2ab=(a+b)^2$ in the equation

$$A-G=\frac{(\sqrt a-\sqrt b)^2}{2}$$

for two positive numbers **a** and **b** $(\sqrt a-\sqrt b)^2$ is always positve or zero, which can be represented as

$$A-G=\frac{(\sqrt a-\sqrt b)^2}{2} \geq 0$$

that implies

$$A \geq G$$

hence proved for two numbers

here $A = G$ would mean that a=b. that is for the same number both AM and GM will be same.

Similarly this can be extrapolated and can be expressed more generally as for $a_1, a_2, a_3, …..a_n$ where all being positive numbers then arithematic mean A = $\frac{a_1 + a_2 + a_3 + ….. + a_n}{n}$ and geometric mean $(a_1a_2a_3…….a_n)^{1/n}$. Then it can be shown that

$$A \geq G$$

and A=G when $a_1= a_2= a_3= …..=a_n$

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