Note that,
If $x,y\to 0$ with $z\to 1$, we have
$$\inf \left\{xy(x+y)^2 + yz(y+z)^2 + zx(z+x)^2\mid x+y+z=1 \wedge x>0\wedge y>0\wedge z>0\right\}=0$$
As for obtaining the inequality $A≥4xyz$, you can easily obtain this result using the Cauchy-Schwarz inequality:
$$(xy(x+y)^2 + yz(y+z)^2 + zx(z+x)^2)\left(\frac 1{xy}+\frac 1{yz}+\frac 1{xz}\right)≥4(x+y+z)^2=4$$
$$\frac A{xyz}≥4\implies A≥4xyz.$$
But, note that $xyz$ is not a constant. Therefore, $4xyz$ is not a "minimum". We only proved that,
$$xy(x+y)^2 + yz(y+z)^2 + zx(z+x)^2≥4xyz$$
The inequality you want to prove.
There is a big error in the last steps of the shorter proof, and a mild gap.
First, the motivation.
Intuitively, this is a question about the so-called taxicab distance. The taxicab distance between $(a,b)$ and $(x,y)$ is $$|x-a|+|y-b|.$$
$|a|+|b|,$ $|a-1|+|b-1|$ and $|a+1|+|b+1|$ are the distances between $(a,b)$ and $(1,1),$ $(0,0),$ and $(-1,-1),$ respectively. And $|a-b|$ is the shortest distance from $(a,b)$ to the line $y=x.$
So everything about this seems to be about the taxicab distances between $(a,b)$ and points $(x,x).$
So it is natural to define the distance from $(a,b)$ to $(x,x)$:
$$f(x)=|a-x|+|b-x|$$
and try to learn about this function.
The big error in the given proof is the last two lines. The statement $b\leq-1\leq a\leq 1$ is wrong.
It should be $$b\leq-1<1\leq a,\tag1$$ and thus $a-b=|a-b|\geq 2.$
The reason for $(1)$ is the skipped step.
$f(x)$ is strictly decreasing when $x<b$ and strictly increasing when $x>a,$ and so the only way three distinct values $u,v,w$ have $f(u)=f(v)=f(w)$ is for the three values to be in $[b,a].$
But we know $f(0)=f(-1)=f(1),$ so $0,1,-1\in[b,a].$
The "shape" of $f$ is the key. We want to show that $f^{-1}(d)$ has at most two points for any $d>|a-b|,$ and for that you need to know not just that $f$ is constant on $[b,a],$ but that the functions is strictly decreasing on $(-\infty,b)$ and strictly increasing on $(a,+\infty).$
So we get a stronger result:
If $u<v<w$ and $f(u)=f(v)=f(w),$ then $|a-b|\geq w-u.$
Picking $a>b$ is possible because the question is symmetric. You could theoretically prove it in two cases, $a>b$ and $b>a,$ but the second case would be exactly the same, with $a,b$ reversed. So we often do only the first case if it is obvious the second case is exactly the same.
Mathematicians will often say things like "Without loss of generality, we can assume $a>b.$" This usually means that we are proving for some subset of cases, but we can easily get the general case back from this subset.
Best Answer
By AM-GM $$3=2\cdot\frac{a}{2}+3\cdot\frac{b}{3}+2\cdot\frac{c}{2}\geq7\sqrt[7]{\left(\frac{a}{2}\right)^2\left(\frac{b}{3}\right)^3\left(\frac{c}{2}\right)^2}.$$ The equality occurs for $\frac{a}{2}=\frac{b}{3}=\frac{c}{2}$ and $a+b+c=3.$
Can you end it now?