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  $x+2y=8$?
 A: $$
x + y + \frac{3}{x} + \frac{9}{2y} = \frac{1}{4}(x+2y) + \left (\frac{3}{4}x + \frac{1}{2}y \right) + \left( \frac{3}{x} + \frac{9}{2y} \right)
= 2 + \left (\frac{3}{4}x + \frac{1}{2}y \right) + \left( \frac{3}{x} + \frac{9}{2y} \right)
$$
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 $2+2*(\sqrt{9}) = 8$
A: 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.
A: 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
$$\frac {x_0} {2}=\frac{\alpha}{x_0} \tag1$$
$$\frac{x_0^2}{3-\alpha} = \frac{(8-x_0)^2}{9}\tag2$$
A: I know this might be a bit stupid, but, I plotted it with:
$$x = (x_1, x_2)$$
$$x \leftarrow x - \vert \nabla^{2} f(x) \vert^{-1} \nabla f(x)$$
And got:
$$y = 3\ \ \ \text{and} \ \ x = 2$$
A: $$x+2y=8  \implies x = 8 - 2y\tag{1}$$
Let $$f(x, y) = x+y+\frac{3}{x}+\frac{9}{2y} \tag{2}$$
Using $(1)$ in $(2)$,
$$f(x, y) = 8 - 2y +y+\frac{3}{8 - 2y}+\frac{9}{2y} = 8 - y +\frac{3}{8 - 2y}+\frac{9}{2y} \tag{3}$$
Differentiate $(3)$ wrt $y$, set it equal to $0$ and solve for $y$
$$\frac{-9}{2y^2} + \frac{3}{2(4-y)^2} - 1 = 0$$
The real root is $y = 3$ and the second derivative $$\left[\frac{9}{y^3} + \frac{3}{(4-y)^3} \right]_{y = 3} = \frac{10}{3} > 0$$
From $(1)$, $x = 2$ and the minimum value is $f(2, 3) = 8$
