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  $x+2y=8$?

The AM-GM inequality, as in the post tag, gives us a lower bound.
Indeed,
$$x+\frac{3}{x} + y +\frac{9}{2y}=2\frac{x+\frac{3}{x}}{2} + 2\frac{y +\frac{9}{2y}}{2}.$$
Since $x,y>0$ we can apply the AM-GM inequality to obtain
$$2\frac{x+\frac{3}{x}}{2} + 2\frac{y +\frac{9}{2y}}{2}\geq 2\sqrt{3}+2\sqrt{\frac{9}{2}}=2(\sqrt 3 + \sqrt\frac{9}{2})\sim 7.71.$$
However, this is just a lower bound.
To find the minimum, we have $$x+y+\frac{3}{x}+\frac{9}{2y}=x+2y-y + \frac{3}{x+2y-2y}+\frac{9}{2y}$$
which is equal to $$8-y +\frac{3}{8-2y}+\frac{9}{2y}.$$
Note that since $x=8-2y>0$, we want $0<y<4$.
A local minimum occurs at $y=3$. 
One can see this by solving for when the derivative of $8-y +\frac{3}{8-2y}+\frac{9}{2y}$ is $0$.
At $y=3$, we have $x=2$ so that $2+3+\frac{3}{2}+\frac{9}{6}=8$ is the minimum desired.
