Inequality trouble: $(a^2+ab+b^2)(b^2+bc+c^2)(c^2+ca+a^2) \geq (ab+bc+ca)^3$ The following inequality is exercise 1.8 from this book.

For any real $a,b,c$, prove the following $$(a^2+ab+b^2)(b^2+bc+c^2)(c^2+ca+a^2) \geq (ab+bc+ca)^3.$$ 

I've managed to prove this via brut-force and Muirhead's inequality (Very unsatisfying). However I'm having difficulty understanding the solution in the back of the book. I'm also interested if anyone else knows of any nice proofs of this inequality?
In particular, the solution at the back says the following (word for word):

If we can show that $\frac{27}{64}(a+b)^2(b+c)^2(c+a)^2\geq (ab+bc+ca)^2$, then the conclusion follows. Denote $S_1=a+b+c$, $S_2=ab+bc+ca$ and $S_3=abc$. We need to show that $27(S_1S_2-S_3)^2\geq 64S_2^3.$

It then goes onto prove the last inequality via cases. However I'm struggling to see how it is sufficient to prove either of those two inequalities. 
We can rewrite $a^2+ab+b^2 = (\frac{a+b}{2})^2+\frac{3}{4}(a^2+b^2)$ and so by AM-GM it is sufficient to prove $$\sqrt{\frac{27}{64}(a+b)^2(b+c)^2(c+a)^2}\sqrt{(a^2+b^2)(b^2+c^2)(c^2+a^2)} \geq  (ab+bc+ca)^3 .$$ However $$\sqrt{(a^2+b^2)(b^2+c^2)(c^2+a^2)} \geq (ab+bc+ca)^2 $$ is clearly wrong and so this does not justify the first claim in the solution. This does give me an idea of where the $\frac{27}{64}$ term comes from however. Other than this I'm a bit stuck and any help would be appreciated. 
Edit: I realised straight after posting that $a^2+ab+b^2 \neq (\frac{a+b}{2})^2+\frac{3}{4}(a^2+b^2)$. So in fact my work is nonsense and can be ignored.
 A: 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
$$
 a^2 + ab + b^2 = \frac 34 (a+b)^2 + \frac 14 (a-b)^2 \ge \frac 34 (a+b)^2
$$
A: 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!
A: I found this identity: $$\text{LHS-RHS} = \frac{1}{6} \sum\, \left( a-b \right) ^{2}\Big[ \left( a+b+c \right) ^{2}{c}^{2}+2\,
 \left( ab+ca+bc \right) ^{2}\Big] \geqq 0$$
