Answer
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Hint: Geometric mean, especially, the square root (such as x) of the product of two numbers (such as a and b) when expressed as the means of a proportion. So, the square of $ y $ is equal to the multiplication of $ x $ and $ z $ respectively.
Complete step by step solution:
If $ y $ is a mean proportion between $ x $ and $ z $ ,
$ \Rightarrow {y^2} = xz $
$ \Rightarrow y = \sqrt {xz} $
We have to prove,
$ {\left[ {\dfrac{{xy + yz + zx}}{{x + y + z}}} \right]^3} = xyz $
So, take numerator of LHS,
$
\Rightarrow {(xy + yz + zx)^3} \\
\Rightarrow {y^3}{(x + y + z)^3} \\
\Rightarrow {y^2}.y{(x + y + z)^3} \\
\Rightarrow zx.y{(x + y + z)^3} \\
\Rightarrow xyz{(x + y + z)^3} \;
$
Now take $ {(x + y + z)^3} $ On LHS,
$ \Rightarrow $ $ {\left[ {\dfrac{{xy + yz + zx}}{{x + y + z}}} \right]^3} = xyz $
Hence proved.
Note:
> The G.M for the given data set is always less than the arithmetic mean for the data set. The ratio of the corresponding observations of the G.M in two series is equal to the ratio of their geometric means.
> If each object in the data set is substituted by the G.M, then the product of the objects remains unchanged.
> The products of the corresponding items of the G.M in two series are equal to the product of their geometric mean.
Complete step by step solution:
If $ y $ is a mean proportion between $ x $ and $ z $ ,
$ \Rightarrow {y^2} = xz $
$ \Rightarrow y = \sqrt {xz} $
We have to prove,
$ {\left[ {\dfrac{{xy + yz + zx}}{{x + y + z}}} \right]^3} = xyz $
So, take numerator of LHS,
$
\Rightarrow {(xy + yz + zx)^3} \\
\Rightarrow {y^3}{(x + y + z)^3} \\
\Rightarrow {y^2}.y{(x + y + z)^3} \\
\Rightarrow zx.y{(x + y + z)^3} \\
\Rightarrow xyz{(x + y + z)^3} \;
$
Now take $ {(x + y + z)^3} $ On LHS,
$ \Rightarrow $ $ {\left[ {\dfrac{{xy + yz + zx}}{{x + y + z}}} \right]^3} = xyz $
Hence proved.
Note:
> The G.M for the given data set is always less than the arithmetic mean for the data set. The ratio of the corresponding observations of the G.M in two series is equal to the ratio of their geometric means.
> If each object in the data set is substituted by the G.M, then the product of the objects remains unchanged.
> The products of the corresponding items of the G.M in two series are equal to the product of their geometric mean.
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