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?


I was wondering if there are alternative approaches to solve this problem

Yes, there is.

and is my proof entirely correct?

I think so. I couldn't find anything wrong with it.

Now let's start with another proof.
Let
$$
\{a,b,c,x,y,z \in \mathbb R_{\geq0} : a=x^{\frac{1}{3}},b=y^{\frac{1}{3}},c=z^{\frac{1}{3}}\}
$$
We know that
$$ 
\begin{equation} 
\begin{aligned}
&a^2+b^2 \geq 2ab \\
\implies &(a+b)(a^2+b^2-ab) \geq ab(a+b) \\
\end{aligned} 
\end{equation} 
$$
$$
a^3+b^3 \geq ab(a+b)
\label{eq1}
\tag{1}
$$
Utilizing this result across $a$, $b$ and $c$ and adding them, we get
$$
2(a^3+b^3+c^3) \geq a^2(b+c)+b^2(c+a)+c^2(a+b)
\label{eq2}
\tag{2}
$$
Now,
$$
a+b \geq 2\sqrt{ab}
$$
Utilizing this result across $a$, $b$ and $c$ and multiplying them, we get
$$
(a+b)(b+c)(c+a) \geq 8abc
$$
On further simplification, we get
$$
a^2(b+c)+b^2(c+a)+c^2(a+b) \geq 6abc
\label{eq3}
\tag{3}
$$
From $\eqref{eq2}$ and $\eqref{eq3}$, it's clear that
$$
2(a^3+b^3+c^3) \geq a^2(b+c)+b^2(c+a)+c^2(a+b) \geq 6abc
$$
Therefore,
$$
\begin{aligned}
&2(a^3+b^3+c^3) \geq 6abc \\
\implies &a^3+b^3+c^3 \geq 3abc \\
\implies &x+y+z \geq 3(xyz)^{\frac{1}{3}}
\end{aligned}
$$
Cubing on both sides, we get
$$
(x+y+z)^3 \geq 27xyz
$$
