I think that there is not a simple proof of this inequality. (for all real numbers)
My Attempt $1$:
Equality occurs for $a=b=c$.
It is quit sure that the inequality is homogeneous.
So normalize with $ab+bc+ca = 1$.
$$
(a^2+b^2+c^2)(ab+bc+ca)^2
\ge
(a^2+2bc)(b^2+2ca)(c^2+2ab)
$$
$$
a^2+b^2+c^2
\ge
(a^2+2bc)(b^2+2ca)(c^2+2ab)
$$
We're done if
$$
1 \ge 2\sum_{cyc}{a^3b^3}+4\sum_{cyc}{a^4bc}+9a^2b^2c^2
$$
By the hypothesis,
$$
(ab+bc+ca)^3 $$$$= a^3 b^3 + 3 a^3 b^2 c + 3 a^3 b c^2 + a^3 c^3 + 3 a^2 b^3 c$$$$ + 6 a^2 b^2 c^2 + 3 a^2 b c^3 + 3 a b^3 c^2 + 3 a b^2 c^3 + b^3 c^3 $$$$= 1= \sum_{cyc}{a^3b^3} + 3\sum_{cyc}{a^3b^2c} + 3\sum_{cyc}{a^3bc^2}+6a^2b^2c^2
$$
$$
3\sum_{cyc}{a^3b^2c} + 3\sum_{cyc}{a^3bc^2}\ge \sum_{cyc}{a^3b^3}+4\sum_{cyc}{a^4bc}+3a^2b^2c^2
$$
As @Michael Rozenberg's feedback in one of my recent previous posts, I have carefully preserved the Equality case for 5 steps. Hope this helps!
I think that this inequality is sharp, as I have already used a blunt inequality. So I think SOS could work.
But this attempt is surely incorrect. Try $a=b=1$ and $c=0$.
My Attempt $2$:
Expand everything.
$$
(a^2+b^2+c^2)(ab+bc+ca)^2
\ge
(a^2+2bc)(b^2+2ca)(c^2+2ab)
$$
Here,
$$
(ab+bc+ca)^2
= a^2b^2+b^2c^2+c^2a^2 + 2abc(a+b+c)
$$
$$
= \sum_{cyc}{a^2b^2}+2abc\sum_{cyc}{a}
$$
Multiplying this thing to $\sum_{cyc}{a^2}$ to get:
$$
= \sum_{cyc}{a^4b^2}+\sum_{cyc}{a^2b^4}+3a^2b^2c^2
$$
$$
+ 2abc (\sum_{cyc}{a^3}+\sum_{cyc}{a^2b}+\sum_{cyc}{ab^2})
$$
As before,
$$
(a^2+2bc)(b^2+2ca)(c^2+2ab)
$$
$$
= 2\sum_{cyc}{a^3b^3}+4abc\sum_{cyc}{a^3}+9a^2b^2c^2
$$
We just need to prove that:
$$
\sum_{cyc}{a^4b^2}+\sum_{cyc}{a^2b^4}+ 2abc (\sum_{cyc}{a^2b}+\sum_{cyc}{ab^2})
$$
$$
\ge 2\sum_{cyc}{a^3b^3}+2abc\sum_{cyc}{a^3}+6a^2b^2c^2
$$
From here, it's very easy to get a wrong inequality again:
As $(4,2)\succ(3,3)$.
$$
\sum_{cyc}{a^4b^2}+\sum_{cyc}{a^2b^4} \ge 2\sum_{cyc}{a^3b^3}
$$
It rests to prove that:
$$
\sum_{cyc}{a^2b}+\sum_{cyc}{ab^2}
\ge \sum_{cyc}{a^3} + 3abc
$$
Supplied
$$
\sum_{cyc}{ab^2} \ge 3abc
$$
The inequality is obviously wrong:
$$
\sum_{cyc}{a^2b} \ge \sum_{cyc}{a^3}
$$
I already have Vask's solution, so I don't need that proof.
I just ask if someone has another good solution, Thanks!
Best Answer
The first, it's easy to check $$(x+a)(x+b)(x+c) = x^3+(a+b+c)x^2+(ab+bc+ca)x+abc.$$ So, setting $X=ab+bc+ca,$ we have $$\prod (a^2+2bc) = \prod \left[X +(a-b)(a-c)\right]$$ $$=X^3+(a^2+b^2+c^2-ab-bc-ca)X^2-(a-b)^2(b-c)^2(c-a)^2$$ $$=X^2\left[(a^2+b^2+c^2-ab-bc-ca)+X\right]-(a-b)^2(b-c)^2(c-a)^2$$ $$=(ab+bc+ca)^2(a^2+b^2+c^2)-(a-b)^2(b-c)^2(c-a)^2$$ $$ \leqslant (ab+bc+ca)^2(a^2+b^2+c^2).$$