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

Here is yet another way, using (weighted) AM-GM.  I would still prefer Holder (the solution posted earlier) for its simplicity, this is just to illustrate it can be done - especially if you know the point of equality.
Let $a = \sqrt{\frac2{15}}$.  Then using weighted AM-GM, we can write the following inequalities.  Note that they are constructed so that the point of equality is easily maintained:
$$5y+2=5a \cdot \frac{y}a+2 \cdot 1 \ge (5a+2)\left(\frac ya \right)^{\frac{5a}{5a+2}}$$
$$\parallel ly, \quad 2z+5=\frac2a \cdot az+5\cdot1 \ge (\tfrac2a+5)(az)^{\frac2{5a+2}}$$
$$x+3y = 1\cdot x + 3a\cdot\frac{y}a \ge (1+3a)x^{\frac1{3a+1}}\left(\frac ya\right)^{\frac{3a}{3a+1}}$$
$$3x+z = 3\cdot x + \frac1a \cdot az \ge (3+\tfrac1a)x^{\frac{3a}{3a+1}}(az)^{\frac1{3a+1}}$$
Multiplying the lot, and substituting the value of $a$, we get
$$\dfrac{(5y+2)(2z+5)(x+3y)(3x+z)}{xyz} \ge 241+44\sqrt{30}$$
