<s>[INST] 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
 [/INST] $$f(x,y,z)=\dfrac{(5y+2)(2z+5)(x+3y)(3x+z)}{xyz}$$
$$\nabla f= \frac{(3 (2+5 y) (5+2 z) (x^2-y z))}{(x^2 y z)} \hat e_x+\frac{(-((2 x-15 y^2) (3 x+z) (5+2 z))}{(x y^2 z))}\hat e_y+\frac{(-((x+3 y) (2+5 y) (15 x-2 z^2))}{(x y z^2))}\hat e_z$$
We get 4 real  solutions:
$$(5/6,-5/18,-5/2),(1,-\sqrt{2/15},\sqrt{15/2}),(1,\sqrt{2/15},\sqrt{15/2}),(6/5,-2/5,-18/5)$$
Out of which only $(1,\sqrt{2/15},\sqrt{15/2})$ is acceptable due to $x,y,z\ge0$
Now the value of function at this point is:
$$f(1,\sqrt{2/15},\sqrt{15/2})=241+44\sqrt{30}$$
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