msm555
Answered

Westonci.ca is the premier destination for reliable answers to your questions, provided by a community of experts. Explore our Q&A platform to find reliable answers from a wide range of experts in different fields. Get immediate and reliable solutions to your questions from a community of experienced professionals on our platform.

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​

Sagot :

Nayefx

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 [tex]\text{$x_1$ and $x_2$}[/tex] where:

  • [tex] x_{1} > x_{2}[/tex]

the AM,GM and HM of [tex]x_1[/tex] and[tex] x_2[/tex] is given by the following table:

[tex]\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}[/tex]

Proof of I:

[tex] \displaystyle \rm AM \times HM = \frac{x_{1} + x_{2}}{2} \times \frac{2}{ \frac{1}{x_{1} } + \frac{1}{x_{2}} } [/tex]

simplify addition:

[tex] \displaystyle \frac{x_{1} + x_{2}}{2} \times \frac{2}{ \dfrac{x_{1} + x_{2}}{x_{1} x_{2}} } [/tex]

reduce fraction:

[tex] \displaystyle x_{1} + x_{2} \times \frac{1}{ \dfrac{x_{1} + x_{2}}{x_{1} x_{2}} } [/tex]

simplify complex fraction:

[tex] \displaystyle x_{1} + x_{2} \times \frac{x_{1} x_{2}}{x_{1} + x_{2}} [/tex]

reduce fraction:

[tex] \displaystyle x_{1} x_{2}[/tex]

rewrite:

[tex] \displaystyle (\sqrt{x_{1} x_{2}} {)}^{2} [/tex]

[tex] \displaystyle AM \times HM = (GM{)}^{2} [/tex]

hence, PROVEN

Proof of II:

[tex] \displaystyle x_{1} > x_{2}[/tex]

square root both sides:

[tex] \displaystyle \sqrt{x_{1} }> \sqrt{ x_{2}}[/tex]

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

[tex]\displaystyle\sqrt{x_{1} } - \sqrt{ x_{2}} > 0[/tex]

square both sides:

[tex]\displaystyle(\sqrt{x_{1} } - \sqrt{ x_{2}} {)}^{2} > 0[/tex]

expand using (a-b)²=a²-2ab+b²:

[tex]\displaystyle x_{1} -2\sqrt{x_{1} }\sqrt{ x_{2}} + x_{2} > 0[/tex]

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

[tex]\displaystyle x_{1} + x_{2} > 2 \sqrt{x_{1} } \sqrt{ x_{2}}[/tex]

divide both sides by 2:

[tex]\displaystyle \frac{x_{1} + x_{2}}{2} > \sqrt{x_{1} x_{2}}[/tex]

[tex]\displaystyle \boxed{ AM>GM}[/tex]

again,

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

expand:

[tex]\displaystyle \frac{1}{x_{1}} - \frac{2}{\sqrt{x_{1} x_{2}} } + \frac{1}{x_{2} }> 0[/tex]

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

[tex]\displaystyle \frac{1}{x_{1}} + \frac{1}{x_{2} }> \frac{2}{\sqrt{x_{1} x_{2}} }[/tex]

[tex]\displaystyle \frac{\frac{1}{x_{1}} + \frac{1}{x_{2} }}{2}> \frac{1}{\sqrt{x_{1} x_{2}} }[/tex]

[tex]\displaystyle \rm \frac{1}{ HM} > \frac{1}{GM} [/tex]

cross multiplication:

[tex]\displaystyle \rm \boxed{ GM >HM}[/tex]

hence,

[tex]\displaystyle \rm A.M>G.M>H.M[/tex]

PROVEN