Inequality trouble: $(a^2+ab+b^2)(b^2+bc+c^2)(c^2+ca+a^2) \geq (ab+bc+ca)^3$

It is a typo. In fact $27(S_1S_2-S_3)^2\geq 64S_2^3$ (which is correct) expands to $$ \frac{27}{64}(a+b)^2(b+c)^2(c+a)^2\geq (ab+bc+ca)^3 $$ with exponent $3$ instead of $2$ on the right-hand side.

Then the wanted inequality follows because $$ a^2 + ab + b^2 = \frac 34 (a+b)^2 + \frac 14 (a-b)^2 \ge \frac 34 (a+b)^2 $$

We use the so called uvw method for proving inequality involving symmetric polynomials of $3$ variables.

Let $a+b+c=3u$, $ab+ac+bc=3v^2$, where $v^2$ can be negative, and $abc=w^3$.

Hence, our inequality it's $kw^6+A(u,v^2)w^3+B(u,v^2)\geq0$, where $A$ and $B$ polynomials such that $A(0,0)=B(0,0)=0$.

Let $\theta^3=1$, where $\theta\neq1$ and $(a,b,c)=(1,\theta,\theta^2)$.

Thus, $a^2+ab+b^2=a^2+ac+c^2=b^2+bc+c^2=u^2=v^2=0$, but $w^3\neq0$,

which says that $k=0$.

Thus, our inequality is a linear inequality of $w^3$,

which says that it's enough to prove our inequality for an extremal value of $w^3$,

which happens for equality case of two variables.

Since our inequality is homogeneous and even degree, we can assume $b=c=1$,

which gives $$3(a^2+a+1)^2\geq(2a+1)^3$$ or $$(a-1)^2(3a^2+4a+2)\geq0,$$ which is obvious.

Done!