$10(x + y) + 4z = 5xyz$. Find the min value of: $A = 2x + y + 2z$ 
PROBLEM: We have $x,y,z \in \mathbb{R}^+$ such as $10(x + y) + 4z = 5xyz$. Find the min value of:
  $$A = 2x + y + 2z$$

My attempt:
I would find the min of:
$$A^2 = \frac{(2x+y+2z)^2(10x+10y+4z)}{5xyz}$$
and then let $z=1$, after that I used Lagrange method to find min of:
$$ \frac{(2x+y+2)^2(10x+10y+4)}{5xy}$$
Finally I found the result which is min value of $A$ when $\dfrac{x}{4}=\dfrac
{y}{6}=\dfrac{z}{5}$
But the problem is, my brother is just a high school student so he can't understand Lagrange's method, so I wonder if having another simple method for this?

For $x=\frac{4}{\sqrt5}$, $b=\frac{6}{\sqrt5}$ and $c=\sqrt5$, we get a value $\frac{24}{\sqrt5}$.
We'll prove that this is a minimal value.
Indeed, let $x=\frac{4a}{\sqrt5}$, $y=\frac{6b}{\sqrt5}$ and $z=\sqrt5c$.
Hence, the condition gives $2a+3b+c=6abc$ and we need to prove that
$$4a+3b+5c\geq12$$ or
$$\frac{(4a+3b+5c)^2(2a+3b+c)}{6abc}\geq144$$ or
$$(4a+3b+5c)^2(2a+3b+c)\geq864abc,$$
which is true by AM-GM:
$$(4a+3b+5c)^2(2a+3b+c)\geq\left(12\sqrt[12]{a^4b^3c^5}\right)^2\cdot6\sqrt[6]{a^2b^3c}=864abc.$$
The equality occurs for $a=b=c=1$.
Done!
