Question

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

Answer 1

Answer:

See below

Step-by-step 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:

$\begin{array}{ |c |c|c | } \hline AM& GM& HM\\ \hline \dfrac{x_{1} + x_{2}}{2} & \sqrt{x_{1} x_{2}} & \dfrac{2}{ \frac{1}{x_{1} } + \frac{1}{x_{2}} } \\ \hline\end{array}$

Proof of I:

$\displaystyle \rm AM \times HM = \frac{x_{1} + x_{2}}{2} \times \frac{2}{ \frac{1}{x_{1} } + \frac{1}{x_{2}} }$

simplify addition:

$\displaystyle \frac{x_{1} + x_{2}}{2} \times \frac{2}{ \dfrac{x_{1} + x_{2}}{x_{1} x_{2}} }$

reduce fraction:

$\displaystyle x_{1} + x_{2} \times \frac{1}{ \dfrac{x_{1} + x_{2}}{x_{1} x_{2}} }$

simplify complex fraction:

$\displaystyle x_{1} + x_{2} \times \frac{x_{1} x_{2}}{x_{1} + x_{2}}$

reduce fraction:

$\displaystyle x_{1} x_{2}$

rewrite:

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

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

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:

$\displaystyle \frac{x_{1} + x_{2}}{2} > \sqrt{x_{1} x_{2}}$

$\displaystyle \boxed{ AM>GM}$

again,

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

expand:

$\displaystyle \frac{1}{x_{1}} - \frac{2}{\sqrt{x_{1} x_{2}} } + \frac{1}{x_{2} }> 0$

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

$\displaystyle \frac{1}{x_{1}} + \frac{1}{x_{2} }> \frac{2}{\sqrt{x_{1} x_{2}} }$

$\displaystyle \frac{\frac{1}{x_{1}} + \frac{1}{x_{2} }}{2}> \frac{1}{\sqrt{x_{1} x_{2}} }$

$\displaystyle \rm \frac{1}{ HM} > \frac{1}{GM}$

cross multiplication:

$\displaystyle \rm \boxed{ GM >HM}$

hence,

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

PROVEN