How find the minimum $\frac{(5y+2)(2z+5)(x+3y)(3x+z)}{xyz}$,if $x,y,z>0$ let $x,y,z&gt;0$, 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
Here is another way, using Holder's inequality:
$$\frac{(5y+2)(2z+5)(x+3y)(3x+z)}{xyz} \geqslant \frac{(\sqrt[4]{5y \cdot 5\cdot x\cdot z}+\sqrt[4]{2\cdot2z\cdot3y\cdot3x})^4}{xyz} \\ = (\sqrt5+\sqrt6)^4=241+44\sqrt{30}$$

with equality iff $\dfrac{5y}2=\dfrac5{2z}=\dfrac{x}{3y}=\dfrac{z}{3x} \iff (x, y, z) = \left(1, \sqrt\frac2{15}, \sqrt\frac{15}2 \right)$.
Hint: 

$5y+2 \geq 2\sqrt{10y}$

$2z+5 \geq 2\sqrt{10z}$

$x+3y \geq 2\sqrt{3xy}$

$3x+z \geq 2\sqrt{3xz}$