<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] Here are two one-line-formula proofs:

*

*By AM-GM:
$$
\frac{x+y+z}{3} \geq \sqrt[3]{x y z} 
$$
Now take the cubic value on both sides of this inequality.


*Use the following identity which also gives you the  exact deviation in positive terms from $27 x y z$: (from which you can derive tighter bounds of the LHS)
$$
(x+y+z)^3 = 27 x y z  + 3 (z-y)^2 x + 3 (x-z)^2 y+ 3 (y-x)^2 z +\\
+ \frac12 (x+y+z)((x-y)^2 + (y-z)^2 + (z-x)^2) 
$$
All terms on the RHS are positive, so you can take lower bounds of the LHS by any term on the RHS or weighted sum of terms on the RHS, with weights between 0 and 1.
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