the maximum value of $xy/(x+2y)$, given $x^2+4y^2+2xy=1, x>0, y>0$

Question

My thought is $\frac{xy}{x+2y} = \frac{1}{2}\frac{x*2y}{x+2y}\le \frac{1}{8}\frac{(x+2y)^2}{x+2y}=\frac{x+2y}{8}$.
I don't know how to proceed.
Please use AM-GM Inequality only.

Answer 1

Sorry I didn't use AM-GM inequatity but solved this problem. From $x^2+4y^2+2xy=1$ we have $(x+y)^2+3y^2=1$. Let $x+y=\cos\theta,y=\sin\theta/\sqrt{3}$, the objective function becomes $\frac{1}{\sqrt{3}}\frac{\cos(\theta+\pi/6)\sin\theta}{\cos(\theta-\pi/6)}$. Let $\varphi=\theta-\pi/6$, then the function is $\frac{1}{2\sqrt{3}}\frac{\cos2\varphi-\sin\pi/6}{\cos\varphi}$ (you can calculate by yourself), which maximizes at $\varphi=0$ and the maximum value is $\frac{1}{4\sqrt{3}}$.

Another method may be near your request.Square the objective function we have $\frac{(xy)^2}{(x+2y)^2}=\frac{t^2}{1+2t}$ where $t=xy$. It's easy to find that the function is monotonely increasing with $t\in(0,\infty)$ while $t\leq \frac{1}{6} $ from the initial condition with AM-GM inequality.

Answer 2

Continuing from OP's work

• Equality in OP's work holds when $ x = 2y$ (which thankfully is the equality case, so they are not off track)

• So we want to maximize $ x + 2y$ given $ x^2 + 2xy + 4y^2 = 1$.

  • If we're allowed to draw the tangent line to the ellipse, then we're quickly done. This tells us the task is not impossible (EG That we haven't over relaxed the inequality)

• Given the condition $x^2 + 2xy + 4y^2 = 1$, we want to maximize $ x+2y$, so let's set $k = x+2y$ and rewrite the condition in terms of $k$.

  • $ 2xy \leq k^2/4$, again with equality when $ x = 2y$.
  • $ (x+2y)^2 = 1 + 2xy \Rightarrow k^2 \leq 1 + k^2/4$.
  • Hence, $ k \leq \frac{2}{\sqrt{3}}$.

• Verify that equality can hold throughout.

  • In particular, what is the equality case $(x, y)$?

• Thus, the maximum to the original inequality is $ \frac{k}{8} \leq \frac{ 1}{ 4 \sqrt{3} } $.