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For positive real numbers \(x,y,\) and \(z,\) find the minimum value of \(\frac{x^3 + 5y^3 + 25z^3}{xyz}.\)

 Jun 3, 2019
 #1
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+1

x=1;y=1;z=1; a=(x^3 + 5*y^3 + 25*z^3) / (x*y*z);printa,y;y++;if(y<100, goto3, discard=0;
The minimum value is when:
x =2 or x = 3, y=1, z=1: (x^3 + 5*y^3 + 25*z^3) / (x*y*z) = 19

 Jun 3, 2019
 #2
avatar+28125 
+3

If we set y = x/51/3 and z = x/251/3 then (x3 + 5y3 + 25z3)/xyz = 15  irrespective of the actual (positive) values of x, y and z.

 Jun 4, 2019

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