As the question states, I am interested in proving the AM GM inequality in 3 variables, namely: $$\sqrt[3]{xyz}\le \frac{(x+y+z)}{3}$$
Clearly, I expanded everything into the following: [a] $$ xyz \le \frac{x^3 + y^3 + z^3 + 3(x^2y + x^2z + y^2x + y^2z + z^2x + z^2y) + 6xyz}{27}$$
I have been given that we know the proof of the AM GM inequality in 2 variables, so I attempted from the base case of the mean between xy and z: [b] $$ \sqrt{(xy)z} \le \frac {xy + z}{2} $$
Thus: [c] $$xyz \le \frac {x^2y^2 +2xyz + z^2}{4}$$
If I can prove that the right hand side of c is less than the right hand side of a, I will have completed my proof. I conjecture that my proof will look like: $$xyz \le \frac {x^2y^2 +2xyz + z^2}{4} \le something \le \frac{x^3 + y^3 + z^3 + 3(x^2y + x^2z + y^2x + y^2z + z^2x + z^2y) + 6xyz}{27}$$
However, this is proving challenging arithmetically. I know that this is possible with calculus, but I am interested in solving the problem without calculus, and from either a geometric, factorial, or arithmetic perspective. Any insight would be appreciated.