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.

  • 2
    $\begingroup$ Try for $2x+y+(x+2y) $. $\endgroup$
    – SarGe
    Aug 10, 2020 at 8:40
  • $\begingroup$ @SarGe You mean @abcd123’s answer, right? But the equality doesn’t hold……. $\endgroup$
    – user808951
    Aug 10, 2020 at 8:44
  • $\begingroup$ Although it seems like you want an elementary solution, I can give you a simple way (more like blowing a canon to kill a mosquito though), use Lagrange multiplier method on $f(x, y) = (x + y)$, subject to the constraint $g(x, y) = xy(x + y) = 2$. :( $\endgroup$ Aug 10, 2020 at 8:49

4 Answers 4


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


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

  • $\begingroup$ There's a small typo on line $3$: the denominator should be $t^2+2t$. $\endgroup$
    – Toby Mak
    Aug 10, 2020 at 9:06
  • 1
    $\begingroup$ @Toby Mak Thank you! I fixed. $\endgroup$ Aug 10, 2020 at 9:09


Let z=x+y,


$(z^2-y^2)y=2 $

$z^2-y^2=2/y $



$\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 $y=1$

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




let $t=\frac{y}{x}>0$


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 $\sqrt3$

  • $\begingroup$ I found @Michael Rozenberg had answered earlier than me lol $\endgroup$ Aug 10, 2020 at 9:03

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