Proof problem involving a inequality

Question

If x,y positive real numbers, then its true that x^3+y^3 ≥ xy(x+y)

Answer 1

We seek to show that x^3 + y^3 >= xy(x+y) = x^2y + xy^2 for x and y positive

x^3 + y^3 =

x^2x + y^2y = (as we want to create the right hand side)

x^2(x+y-y) + y^2(y+x-x) =

x^2y + x^2(x-y) + xy^2 + y^2(y-x) =

x^2y + xy^2 + (x^2-y^2)(x-y) =

x^2y + xy^2 + (x-y)(x+y)(x-y) =

x^2y + xy^2 + (x-y)^2(x+y)

As x and y are positive, x + y > 0, and (x-y)^2 >= 0 for all x and y. Thus,

x^2y + xy^2 + (x-y)^2(x+y) >= x^2y + xy^2

Thus,

x^3 + y^3 >= x^2y + xy^2