For $ x,y,z $ positive real numbers satisfying 2$k^3$ + (x+y+z)$k^2$ - xyz = 0 where k is positive integer. Prove that xyz ≥ $(2k)^3$

Question

Let $x,y,z $ be positive real numbers satisfying

2$k^3$ + (x+y+z)$k^2$ - xyz = 0 where $k $ is a positive integer.

Prove that xyz ≥ $(2k)^3$ .

Apart from using $Lagrange $ multiplier, any other method to prove that the minimum is for $x=y=z=2k$ .

Answer 1

Hint: Write your equation in the form $$\frac{2k^3-xyz}{k^2}=x+y+z\geq 3\sqrt[3]{xyz}$$ then you will get an inequality in $$xyz$$