Using expansion (x+ y + z)3, show that (x + y + z)3 >= 27xyz
I've written out expansion but cannot solve question. Any help appreciated.
\begin{align*}&\quad (x+y+z)^3\\&= x^3+y^3+z^3 + 3(x^2y+x^2z+y^2x+y^2z+z^2x+z^2y) + 6xyz\\ &= (x^3+y^3+z^3) + 3z(x^2+y^2) + 3y(x^2+z^2) + 3x(y^2+z^2) + 6xyz\\ &\ge (x^3+y^3+z^3) + 6zxy + 6yxz + 6xyz + 6xyz\\ &= x^3+y^3+z^3 + 24xyz \end{align*}
This is really just the AM-GM inequality with three variables. The AM-GM inequality with four variables is easy to prove, and
you can use that one to prove the version with three variables.
To provide some intuition: Since we have to expand a cube involving three terms to prove something with \(xyz\) in the RHS, it was already extremely likely that \( a+b+c\ge 3\sqrt[3]{abc} \) would be useful (by simply letting \(a = x^3\) and etc.). But this is quite tedious to prove, and there's still no guarantee that it'd do all the magic. So I started by considering every other term.
The terms \(x^2y + xy^2 + x^2z+xz^2+y^2z+yz^2 \) all took the form something-squared plus something. Just like in the above case, \(a+b \ge 2\sqrt{ab} \) could have been really useful. But I didn't want renegade square roots appearing because there's no square root at all in what I'm trying to prove.
Since a square and square root cancel each other out when dealing with positive numbers, it made sense to apply that on the squares instead. Hence, I used \(x^2+y^2\ge 2xy \) instead of \( x+y \ge 2\sqrt{xy} \).
But of course, I had to rearrange the terms and do some factorising before I could even think of using that formula.
It was from there, I saw that I suddenly had almost all of the \(xyz\) in place (I already had 24 of them, and only needed 3 more). Well, the last 3 takes effort to get at, but at least I know I could get there