What is the minimum value of $x+y$?

Question

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

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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.

Answer 1

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 $x=\sqrt{3}-1$.

Answer 2

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 $$2(t+1)^3-3\sqrt3(t^2+2t)=(t-\sqrt3+1)^2(2t+2+\sqrt3)\geq0.$$

Answer 3

$xy(x+2y)=2$.

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 $y=1$

$z^2=2/1+1=3$, $z=3^{0.5}$, $x=3^{0.5}-1$,

$x+y=3^{0.5}$

Answer 4

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 $\sqrt3$