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  $x+2y=8$?

$$
x + y + \frac{3}{x} + \frac{9}{2y} = \frac{1}{4}(x+2y) + \left (\frac{3}{4}x + \frac{1}{2}y \right) + \left( \frac{3}{x} + \frac{9}{2y} \right)
= 2 + \left (\frac{3}{4}x + \frac{1}{2}y \right) + \left( \frac{3}{x} + \frac{9}{2y} \right)
$$
And
$$
\left (\frac{3}{4}x + \frac{1}{2}y \right) + \left( \frac{3}{x} + \frac{9}{2y} \right) \geq 2 \sqrt{\left (\frac{3}{4}x + \frac{1}{2}y \right) \left( \frac{3}{x} + \frac{9}{2y} \right)}
$$ (AM-GM, equality: $\left (\frac{3}{4}x + \frac{1}{2}y \right) = \left( \frac{3}{x} + \frac{9}{2y} \right)$)
Further,
$$
\left (\frac{3}{4}x + \frac{1}{2}y \right) \left( \frac{3}{x} + \frac{9}{2y} \right) \geq \left( \frac{3}{2} + \frac{3}{2} \right)^2 = 9
$$ (Cauchy– Schwarz inequality, equality: $\frac{4}{x^2} = \frac{9}{y^2}$)
Observe that the equalities hold when $x=2$ and $y=3$, and the minimum is $2+2*(\sqrt{9}) = 8$
