if x and y are positive numbers such that xy^2=1 prove that x+2y >= 3

Answer 1

xy² = 1 ⇒ x = 1/y²

Write x+2y as 1/y² + 2y

Let w(y) = 1/y² + 2y

Take derivative of w wrt y, set to 0 to find critical point:

dw/dy = -2/y³ + 2

dw/dy = 0 = -2/y³ + 2 ⇒ y = 1

d²w/dy² = 6/y³ @ y =1 this is positive, hence the function is concave up and the critical value is a minimum value.

w(1) = 1/1² + 2*1 = 3 <<< therefore w(y) ≥ 3 and (for xy² = 1), x+2y ≥ 3