Determine extreme values of bcx+cay+abz subject to conditions xyz=abc where a>0, b>0,c>0

Question

Using lagrange multiplier i came to know the stationary point are (a,b,c) and putting back it in equation i got abc+abc+abc = 3abc. But after that i dont know how to proceed further and how to conclude the answer. My professor told me that i can use the concept AM>GM, but I don't know how to use this concept here?

Answer 1

Hint (assuming $x,y,z \gt 0$ since your professor mentioned AM-GM):

$$ \frac{bcx+cay+abz}{3} \ge \sqrt[3]{a^2b^2c^2xyz} = \sqrt[3]{a^3b^3c^3} = abc $$

As for the upper bound, consider the case where $y,z \to 0\,$.