# 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.

• Try for $2x+y+(x+2y)$. Aug 10, 2020 at 8:40
• @SarGe You mean @abcd123’s answer, right? But the equality doesn’t hold…….
– user808951
Aug 10, 2020 at 8:44
• 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 $g(x, y) = xy(x + y) = 2$. :( Aug 10, 2020 at 8:49

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

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

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

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 $$y=1$$

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

$$x+y=3^{0.5}$$

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 $$\sqrt3$$

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