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?

My favorite technique for proving symmetric inequalities of positive numbers (particularly if you have a computer algebra package) is to note that if the inequality is symmetric, then w.l.o.g. we can assume the variables are in sorted order, then rewrite the inequality using the smallest variable and the consecutive differences, expand everything algebraically and note that all the coefficients are positive.
Using the example at hand
$(x + y + z)^3 - 27 x y z \ge 0$
assume w.l.o.g. $x\le y \le z$ and let $y=x+a$ and $z = x + a + b$, so
$\begin{align*}(x + y + z)^3 - 27 x y z &= (3x + 2a + b)^3 - 27 x (x+a)(x+a+b) \\ &= 9 a b x + 6 a b^2 + 9 x a^2 + 9 x b^2 + 12 b a^2 + b^3 + 8a^3\end{align*}$
which is greater than or equal to $0$ as all of $x$, $a$, and $b$ are.
This trick does not always work, but it works surprisingly often.
