SOLUTION: Let x, y, z be positive real numbers such that {{{xyz = 8.}}} Find the minimum value of {{{x + 2y + 4z.}}}

Algebra ->  Inequalities -> SOLUTION: Let x, y, z be positive real numbers such that {{{xyz = 8.}}} Find the minimum value of {{{x + 2y + 4z.}}}      Log On


   



Question 1156957: Let x, y, z be positive real numbers such that xyz+=+8. Find the minimum value of x+%2B+2y+%2B+4z.
Answer by ikleyn(52800) About Me  (Show Source):
You can put this solution on YOUR website!
.

In Math, there is well known Arithmetic Mean - Geometric Mean inequality, which for 3 variables has the form



    for any positive real numbers a, b and c   %28a%2Bb%2Bc%29%2F3 >= root%283%2Ca%2Ab%2Ac%29,

    and inequality becomes an EQUALITY at a = b = c, ONLY.



Apply it to a = x, b = 2y, c = 4z. You will get


    %28x+%2B+2y+%2B+4z%29%2F3 >= root%283%2C+%28x%2A2y%2A4z%29%29 = root%283%2C8xyz%29 = root%283%2C8%2A8%29 = root%283%2C64%29 = 4.


Thus  x + 2y + 4z >= 3*4 = 12.


Again, x + 2y + 4z >= 12  for all positive real x, y and z, for which xyz = 8.


Since the equality is ACHIEVED at x = 2y = 4z, the minimum of x + 2y + 4z is 12.   ANSWER

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On Arithmetic Mean - Geometric Mean inequality see this Wikipedia article

https://en.wikipedia.org/wiki/Inequality_of_arithmetic_and_geometric_means