How to approach this algebraic problem

Question

I have no idea how to approach this problem apart from expanding the whole thing out which creates more mess. And I am sure there is a easy trick. I think it is because I lack experience on such problems to spot the pattern.

Answer 1

If $x,y,z>0$, then AM-GM inequality says $$\frac{x+y+z}{3} \geq \sqrt[3]{xyz}.$$ Let $x=a^2b, y=b^2c, z=c^2a$, then you get $$\frac{a^2b+b^2c+c^2a}{3} \geq \sqrt[3]{a^3b^3c^3}=abc$$ Now try with the other expression.