What is the minimum value of $x+y$? 
Suppose $x,y$ are positive real numbers that satisfy $$xy(x+2y)=2$$ What is the minimum value of $x+y$?


My Thoughts
I’ve attempted using arithmetic-geometric mean inequality and got:
$\frac{x+y+x+2y}{3} \geq \sqrt[3]{2}$
Therefore $2(x+y)+y \geq 3\sqrt[3]{2}$, then I got trapped.
Feels like I’m in the wrong way, I need a hint.

Let $k$ be a minimal value of $x+y$.
Thus, $$x+y\geq k$$ or
$$\frac{2(x+y)^3}{xy(x+2y)}\geq k^3$$ or for $x=ty$
$$\frac{2(t+1)^3}{t^2+2t}\geq k^3$$ and since $$\min_{t>0}\frac{2(t+1)^3}{t^2+2t}=3\sqrt3,$$ which  occurs for $t=\sqrt3-1,$ we obtain that $k=\sqrt3$.
Can you get now a full solution?
By the way, we can get the last result without derivatives because
$$2(t+1)^3-3\sqrt3(t^2+2t)=(t-\sqrt3+1)^2(2t+2+\sqrt3)\geq0.$$
