Question
Answers

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)} $.

Answer Verified Verified
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. \[\dfrac{{x + y}}{{\text{2}}} \geqslant \sqrt {xy} \].
Bookmark added to your notes.
View Notes
×