Inequality: $(x + y + z)^3 \geq 27 xyz$ Edit: $a,b,c$ and $x,y,z$ are positive, real numbers. 
Since $(a-b)^2 \geq 0~$, $a^2 + b^2 - 2ab\geq0~$ and $a^2 + b^2 \geq 2ab~$. Similarly, $a^2 + c^2 \geq 2ac~$ and $b^2 + c^2 \geq 2bc~$.  
Adding these inequalities together, $2(a^2 +b^2 + c^2) \geq 2(ab + ac +bc)~$ and accordingly, $a^2 +b^2 + c^2 \geq ab + ac +bc~$  
Multiplying both sides by $(a + b + c)$:   
$(a^2 +b^2 + c^2)(a+b +c) \geq (ab + ac +bc)(a + b + c)~~$ and simplifying this, $ a^3 + b^3 + c^3 + \Sigma a^2 b \geq 3abc + \Sigma a^2 b $
Therefore, it follows that $a^3 + b^3 + c^3 \geq 3abc~$, and letting $a^3 = x~$, $b^3 = y~$, $c^3 = z~$: $x + y + z\geq 3\sqrt[3]{xyz}$   
Cubing both sides, $(x + y + z)^3 \geq 27 xyz~$ which was to be proven.
I was wondering if there are alternative approaches to solve this problem (possibly using higher-level maths), and is my proof entirely correct?

The following uses what is perhaps the most practical formulation of the AM-GM inequalities, which can be found as Theorem 2.6a in Ivan Niven's excellent Maxima and Minima without Calculus. 

Theorem 2.6a If $n$ positive functions have a fixed product, their sum is minimum if it can be arranged that the functions are equal. On the other hand, if $n$ positive functions have a fixed sum, their product is maximum if it can be arranged that the functions are equal.

By "positive" functions, Niven means functions that are positive on the domain we care about.
To see how this applies to the problem at hand, we see that it is enough to show that $(x+y+z)^3\geq 27k$ where $k$ is the product of $xyz$. Evidently, the theorem applies as the product is constant, so the minimum on the left-hand side is given by when $x=y=z=\sqrt[3]{k}$, and hence the minimum of the left-hand side is in fact $27k$ as desired.

The statement and (one) proof of the AM-GM inequalities can be found on page 21 of Niven's book, while the proof of Theorem 2.6a begins at the bottom of page 27.
