How find the minimum $\frac{(5y+2)(2z+5)(x+3y)(3x+z)}{xyz}$,if $x,y,z>0$ 
let $x,y,z>0$, find the minimum of the value
  $$\dfrac{(5y+2)(2z+5)(x+3y)(3x+z)}{xyz}$$

I think we can use AM-GM inequality to find it.
$$5y+2=y+y+y+y+y+1+1\ge 7\sqrt[7]{y^5}$$
$$2x+5=x+x+1+1+1+1+1\ge 7\sqrt[7]{x^2}$$
$$x+3y=x+y+y+y\ge 4\sqrt[4]{xy^3}$$
$$3x+z=x+x+x+z\ge 4\sqrt[4]{x^3z}$$
but this is not true,because not all four equalities can hold at once.
This problem is from china middle school test,so I think have without Lagrange methods,so I think this inequality have AM-GM inequality

Hint: 
$5y+2 \geq 2\sqrt{10y}$
$2z+5 \geq 2\sqrt{10z}$
$x+3y \geq 2\sqrt{3xy}$
$3x+z \geq 2\sqrt{3xz}$
