If $x+2y=8$, find the minumum of $x+y+\frac{3}{x}+\frac{9}{2y}$ (x, y ∈ R+) Question : If $x+2y=8$, find the minumum of $x+y+\dfrac{3}{x}+\dfrac{9}{2y}$ $(x, y \in \mathbb{R}^+)$
I tried to use $x=8-2y$, and I thought I can plug in. But I think I'm not doing it well. I thought about AM-GM, and how can I expand this  using  $x+2y=8$?

Knowing the answer from taking the derivative, I'd cheat by "observing"
$$x+y+\frac 3x + \frac{9}{2y}=\left(\frac x2+y\right) + \left(\frac x2 + \frac 2x\right)+\left(\frac 1x+\frac {9}{2y}\right)$$
$$\ge 4+2\sqrt{\frac x2 \cdot \frac 2x}+\frac{(x+2y)(\frac 1x + \frac{9}{2y})}{x+2y} \tag{AM-GM}$$
$$\ge 4+2+\frac{(1+3)^2}{8}=8 \tag{C-S}$$
Without knowing the answer and barred from using calculus, I'd do
$$x+y+\frac 3x + \frac{9}{2y}=\left(\frac x2+y\right) + \left(\frac x2 + \frac \alpha x\right)+\left(\frac {3-\alpha}{ x}+\frac {9}{2y}\right)\\
$$
and solve for $\alpha$ and $x_0$ such that
$$\frac {x_0} {2}=\frac{\alpha}{x_0} \tag1$$
$$\frac{x_0^2}{3-\alpha} = \frac{(8-x_0)^2}{9}\tag2$$
