# Let the arithmetic & geometric mean of two numbers be \[A\] and $G$ respectively, then prove that the numbers are $A \pm \sqrt {\left( {A + G} \right)\left( {A - G} \right)} $.

Last updated date: 27th Mar 2023

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Answer

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Hint: Take any two variables, find their A.M (Arithmetic Mean) and G.M (Geometric Mean) accordingly and then simplify them.

Complete step-by-step answer:

Let $x$ and $y$ be the two numbers whose A.M is $A$ and G.M is $G$ respectively

i.e. $x = A + \sqrt {\left( {A + G} \right)\left( {A - G} \right)} $

$y = A - \sqrt {\left( {A + G} \right)\left( {A - G} \right)} $

where \[A = \dfrac{{x + y}}{2}\] and \[G = \sqrt {xy} \]

Now solving $A \pm \sqrt {\left( {A + G} \right)\left( {A - G} \right)} $

$ = \left( {\dfrac{{x + y}}{2}} \right) \pm \sqrt {{{\left( {\dfrac{{x + y}}{{\text{2}}}} \right)}^2} - {{\left( {\sqrt {xy} } \right)}^2}} $

Expanding terms inside the square root

\[

\\

= \left( {\dfrac{{x + y}}{2}} \right) \pm \sqrt {\left( {\dfrac{{{x^2} + {y^2} + 2xy}}{4}} \right) - \left( {xy} \right)} \\

= \left( {\dfrac{{x + y}}{2}} \right) \pm \sqrt {\dfrac{{{x^2} + {y^2} + 2xy - 4xy}}{4}} \\

= \left( {\dfrac{{x + y}}{{\text{2}}}} \right) \pm \sqrt {\dfrac{{{x^2} + {y^2} - 2xy}}{4}} \\

= \left( {\dfrac{{x + y}}{{\text{2}}}} \right) \pm \sqrt {{{\left( {\dfrac{{x - y}}{{\text{2}}}} \right)}^2}} \\

= \left( {\dfrac{{x + y}}{2}} \right) \pm \left( {\dfrac{{x - y}}{{\text{2}}}} \right) \\

\]

First let’s consider the positive sign, we get

\[

= \left( {\dfrac{{x + y}}{{\text{2}}}} \right) + \left( {\dfrac{{x - y}}{{\text{2}}}} \right) \\

= \dfrac{x}{2} + \dfrac{y}{2} + \dfrac{x}{2} - \dfrac{y}{2} \\

= \dfrac{x}{2} + \dfrac{x}{2} \\

= x \\

\]

Now let’s consider the negative sign, we get

\[

= \left( {\dfrac{{x + y}}{{\text{2}}}} \right) - \left( {\dfrac{{x - y}}{{\text{2}}}} \right) \\

= \dfrac{x}{2} + \dfrac{y}{2} - \dfrac{x}{2} + \dfrac{y}{2} \\

= \dfrac{y}{2} + \dfrac{y}{2} \\

= y \\

\]

Thus, \[x = A + \sqrt {\left( {A + G} \right)(A - G)} \]

\[y = A - \sqrt {\left( {A + G} \right)(A - G)} \]

Hence proved.

Note: In these types of problems, assume required variables to arrive at the solution. Always remember that the Arithmetic Mean of a list of non-negative real numbers is greater than or equal to the Geometric Mean of the same list i.e. \[\dfrac{{x + y}}{{\text{2}}} \geqslant \sqrt {xy} \].

Complete step-by-step answer:

Let $x$ and $y$ be the two numbers whose A.M is $A$ and G.M is $G$ respectively

i.e. $x = A + \sqrt {\left( {A + G} \right)\left( {A - G} \right)} $

$y = A - \sqrt {\left( {A + G} \right)\left( {A - G} \right)} $

where \[A = \dfrac{{x + y}}{2}\] and \[G = \sqrt {xy} \]

Now solving $A \pm \sqrt {\left( {A + G} \right)\left( {A - G} \right)} $

$ = \left( {\dfrac{{x + y}}{2}} \right) \pm \sqrt {{{\left( {\dfrac{{x + y}}{{\text{2}}}} \right)}^2} - {{\left( {\sqrt {xy} } \right)}^2}} $

Expanding terms inside the square root

\[

\\

= \left( {\dfrac{{x + y}}{2}} \right) \pm \sqrt {\left( {\dfrac{{{x^2} + {y^2} + 2xy}}{4}} \right) - \left( {xy} \right)} \\

= \left( {\dfrac{{x + y}}{2}} \right) \pm \sqrt {\dfrac{{{x^2} + {y^2} + 2xy - 4xy}}{4}} \\

= \left( {\dfrac{{x + y}}{{\text{2}}}} \right) \pm \sqrt {\dfrac{{{x^2} + {y^2} - 2xy}}{4}} \\

= \left( {\dfrac{{x + y}}{{\text{2}}}} \right) \pm \sqrt {{{\left( {\dfrac{{x - y}}{{\text{2}}}} \right)}^2}} \\

= \left( {\dfrac{{x + y}}{2}} \right) \pm \left( {\dfrac{{x - y}}{{\text{2}}}} \right) \\

\]

First let’s consider the positive sign, we get

\[

= \left( {\dfrac{{x + y}}{{\text{2}}}} \right) + \left( {\dfrac{{x - y}}{{\text{2}}}} \right) \\

= \dfrac{x}{2} + \dfrac{y}{2} + \dfrac{x}{2} - \dfrac{y}{2} \\

= \dfrac{x}{2} + \dfrac{x}{2} \\

= x \\

\]

Now let’s consider the negative sign, we get

\[

= \left( {\dfrac{{x + y}}{{\text{2}}}} \right) - \left( {\dfrac{{x - y}}{{\text{2}}}} \right) \\

= \dfrac{x}{2} + \dfrac{y}{2} - \dfrac{x}{2} + \dfrac{y}{2} \\

= \dfrac{y}{2} + \dfrac{y}{2} \\

= y \\

\]

Thus, \[x = A + \sqrt {\left( {A + G} \right)(A - G)} \]

\[y = A - \sqrt {\left( {A + G} \right)(A - G)} \]

Hence proved.

Note: In these types of problems, assume required variables to arrive at the solution. Always remember that the Arithmetic Mean of a list of non-negative real numbers is greater than or equal to the Geometric Mean of the same list i.e. \[\dfrac{{x + y}}{{\text{2}}} \geqslant \sqrt {xy} \].

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