Question

Let x, y, z be positive real no such that x + y + z = 1. What is the minimum value of 1x+4y+9z1x+4y+9z is

Answer 1

Step-by-step explanation:

Given Let x, y, z be positive real no such that x + y + z = 1. What is the minimum value of 1 x+4 y+9 z is

• If x + y + z = 1, the maximum occurs when x = y = z = 1/3.
• Therefore x + 4 y + 9 z will be
•               1/3 + 4 (1/3) + 9 (1/3)
•             = 1/3 + 4/3 + 3
•             = 1 + 4 + 9 / 3
•             = 14/3 is the value.
• So for all x,y,z ϵ R, such that x + y + z = 1

Then x + 4y + 9z  ≤  14/3