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.
 A: Let $t=x+y$ and we need $t_{\min}$. Then we have $$(t+y)(t-y)y=2\implies t^2y-y^3=2$$ and thus $$t^2 = y^2+{2\over y}$$ so if we apply Am-Gm for three terms we get $$t^2=y^2+{1\over y}+{1\over y}\geq 3$$
and minimum value $t=\sqrt{3}$ is achieved iff $y^2 = {1\over y}$ i.e. $y=1$ and $x=\sqrt{3}-1$.
A: $xy(x+2y)=2$.
Let z=x+y,
$(z-y)y(z+y)=2$
$(z^2-y^2)y=2 $
$z^2-y^2=2/y $
$z^2=2/y+y^2$
$z=(2/y+y^2)^{0.5}$
$\frac {d}{dx} (2/y-y^2)^{0.5}=\frac{y^3-1}{y^2((y^3+2)/y)^{0.5}} $which equals $0$ at $y=1$
$z^2=2/1+1=3$, $z=3^{0.5}$, $x=3^{0.5}-1$,
$x+y=3^{0.5}$
A: 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
$$2(t+1)^3-3\sqrt3(t^2+2t)=(t-\sqrt3+1)^2(2t+2+\sqrt3)\geq0.$$
A: homogeneous.
let $t=\frac{y}{x}>0$
$\frac{(x+y)^3}{xy(x+2y)}=\frac{(t+1)^3}{t(2t+1)}$
using derivative we know its minimum is $\frac{3\sqrt3}{2}$,
at $t=(1+\sqrt3){2}$
so from, $\begin{matrix}{\frac{y}{x}=(1+\sqrt3)/2\\xy(x+2y)=2}\end{matrix}$
we can actually solve an $(x,y)$
So the answer is $\sqrt3$
