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?
 A: Look up "arithmetic-geometric mean inequality".  Your proof is fine, if you assume the variables $\ge 0$, except that your notation $\Sigma a^2 b$ is nonstandard.
A: 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.
A: I know a nice proof. It goes like this:
Let $x,y,z>0$. You know that $\frac{x+y}{2} \geq \sqrt{xy}$. This can be generalized for four numbers
$$\frac{a+b+c+d}{4}=\frac{\frac{a+b}{2}+\frac{c+d}{2}}{2}\geq \sqrt[4]{abcd}.$$
Now pick $a=x,b=y,c=z,d=\sqrt[3]{xyz}$ and you'll get your inequality.
For $x,y,z$ not positive the inequality may not hold. Check $x=-1, y=-2, z=-3$.
A: 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.
A: 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
A: 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.
A: Since the inequality is homogeneous we may WLOG assume that $xyz=1$ and we have  to prove the inequality $(x+y+z)^3 \geq 27$ or $x+y+z \geq 3.$  But the last inequality we obtain immediately from $AM \geq GM$ under the assumption $xyz=1.$
A: 
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
$$
(x+y+z)^3 \geq 27xyz
$$
