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Prove that A.M, G.M. and H.M between any two unequal positive numbers satisfy the following relations.

i. (G.M)²= (A.M)×(H.M)
ii.A.M>G.M>H.M​

User Ants Aasma
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Answer:

See below

Explanation:

we want to prove that A.M, G.M. and H.M between any two unequal positive numbers satisfy the following relations.

  1. (G.M)²= (A.M)×(H.M)
  2. A.M>G.M>H.M

well, to do so let the two unequal positive numbers be
\text{$x_1$ and $x_2$} where:


  • x_(1) > x_(2)

the AM,GM and HM of
x_1 and
x_2 is given by the following table:


\begin{array}  \hline AM& GM& HM\\ \hline (x_(1) + x_(2))/(2) & \sqrt{x_(1) x_(2)} & (2)/( (1)/(x_(1) ) + (1)/(x_(2)) ) \\ \hline\end{array}

Proof of I:


\displaystyle \rm AM * HM = (x_(1) + x_(2))/(2) * (2)/( (1)/(x_(1) ) + (1)/(x_(2)) )

simplify addition:


\displaystyle (x_(1) + x_(2))/(2) * (2)/( (x_(1) + x_(2))/(x_(1) x_(2)) )

reduce fraction:


\displaystyle x_(1) + x_(2) * (1)/( (x_(1) + x_(2))/(x_(1) x_(2)) )

simplify complex fraction:


\displaystyle x_(1) + x_(2) * (x_(1) x_(2))/(x_(1) + x_(2))

reduce fraction:


\displaystyle x_(1) x_(2)

rewrite:


\displaystyle (\sqrt{x_(1) x_(2)} {)}^(2)


\displaystyle AM * HM = (GM{)}^(2)

hence, PROVEN

Proof of II:


\displaystyle x_(1) > x_(2)

square root both sides:


\displaystyle \sqrt{x_(1) }> \sqrt{ x_(2)}

isolate right hand side expression to left hand side and change its sign:


\displaystyle\sqrt{x_(1) } - \sqrt{ x_(2)} > 0

square both sides:


\displaystyle(\sqrt{x_(1) } - \sqrt{ x_(2)} {)}^(2) > 0

expand using (a-b)²=a²-2ab+b²:


\displaystyle x_(1) -2\sqrt{x_(1) }\sqrt{ x_(2)} + x_(2) > 0

move -2√x_1√x_2 to right hand side and change its sign:


\displaystyle x_(1) + x_(2) > 2 \sqrt{x_(1) } \sqrt{ x_(2)}

divide both sides by 2:


\displaystyle (x_(1) + x_(2))/(2) > \sqrt{x_(1) x_(2)}


\displaystyle \boxed{ AM>GM}

again,


\displaystyle \bigg( \frac{1}{\sqrt{x_(1) }} - \frac{1}{\sqrt{ x_(2)}} { \bigg)}^(2) > 0

expand:


\displaystyle (1)/(x_(1)) - \frac{2}{\sqrt{x_(1) x_(2)} } + (1)/(x_(2) )> 0

move the middle expression to right hand side and change its sign:


\displaystyle (1)/(x_(1)) + (1)/(x_(2) )> \frac{2}{\sqrt{x_(1) x_(2)} }


\displaystyle ((1)/(x_(1)) + (1)/(x_(2) ))/(2)> \frac{1}{\sqrt{x_(1) x_(2)} }


\displaystyle \rm (1)/( HM) > (1)/(GM)

cross multiplication:


\displaystyle \rm \boxed{ GM >HM}

hence,


\displaystyle \rm A.M>G.M>H.M

PROVEN

User Jim Green
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