Positive Triplets Equality
If $ x, $ $ y, $ and $ z $ are positive real numbers such that \[\frac{x + 2y + 4z + 8}{4} = \frac{4}{\frac{1}{x} + \frac{1}{2y} + \frac{1}{4z} + \frac{1}{8}}.\]Find the ordered triple $ (x,y,z) $.
- 1
- 2
- 3
- +
- 4
- 5
- 6
- -
- 7
- 8
- 9
- $\frac{a}{b}$
- .
- 0
- =
- %
- $a^n$
- $a^{\circ}$
- $a_n$
- $\sqrt{}$
- $\sqrt[n]{}$
- $\pi$
- $\ln{}$
- $\log$
- $\theta$
- $\sin{}$
- $\cos{}$
- $\tan{}$
- $($
- $)$
- $[$
- $]$
- $\cap$
- $\cup$
- $,$
- $\infty$