<s>[INST] 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?
 [/INST] An elementary approach, without $\text{AM} \ge \text{GM}$ is to use the identity
$$x^3 + y^3 + z^3 - 3xyz = (x+y+z)\left(\frac{(x-y)^2 + (y-z)^2 + (z-x)^2}{2}\right)$$
Thus $$\text{if } \sqrt[3]{a} + \sqrt[3]{b} + \sqrt[3]{c} \ge 0\  \text{then } (a+b+c)^3 \ge 27abc$$
by setting $x = \sqrt[3]{a}$ etc
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