Cyclic inequality with $a+b+c=1$ 
Let $a;b;c$ be positive real numbers such that $a+b+c=1$
Prove: $$a^2+b^2+c^2+a^2b+b^2c+c^2a+4abc \geq \dfrac{16}{27}$$


First, I tried with $abc \geq (a+b-c)(b+c-a)(c+a-b)$
But it gave nothing.
Then I tried to add: $a^2b+\dfrac{b}{9}+b^2c+\dfrac{c}{9}+c^2a+\dfrac{a}{9}-\dfrac{1}{9}$
Then AM-GM, but from it, I got: $\dfrac{2}{3}(a^2+b^2+c^2)+4abc \geq \dfrac{10}{27}$
I think it's wrong, I'm stuck here.
Can someone help? Thanks.
 A: Claim:$$27(a^2c+b^2a+c^2b+abc)\le 4{(a+b+c)}^3$$
Proof:WLOG $a=\min(a,b,c)$ then take $b=a+u,c=a+w$ where $u,w\ge 0$ then we have to prove $$-9au^2+9auw-9aw^2-4u^2+15u^2w-12uw^2-4w^3\le 0$$ $$-9a(u^2-uw+w^2)-{(u-2w)}^2(4u+w)\le 0$$ which is true

thus $$a^2c+b^2a+c^2b+abc\le \frac{4}{27}\tag 1$$
we have to prove  $$a^2+b^2+c^2+4abc+a^2b+b^2c+c^2a\ge \frac{16}{27}$$ using (1) it suffices to prove:$$a^2+b^2+c^2+ab(a+b)+bc(b+c)+ac(a+c)+5abc\ge \frac{16}{27}+\frac{4}{27}$$ using $a+b+c=1$ and some mainipulation we have to prove $$ab+bc+ca-2abc\le \frac{7}{27}$$ WLOG $c=\min(a,b,c)$ then $c\le 1/3$ or $1-2c>0$ $$ab+bc+ca-2abc-\frac{7}{27}=\color{red}{ab}(1-2c)+c(1-c)-\frac{7}{27}$$ $$\le \color{red}{\frac{{(a+b)}^2}{4}}(1-2c)+c(1-c)-\frac{7}{27}\tag{AM-GM}$$ $$=\frac{{(1-c)}^2}{4}(1-2c)+c(1-c)-\frac{7}{27}$$ $$={(3c-1)}^2(6c+1)\frac{-1}{108}\le 0$$ which is true and proof is completed(phew!)
Acknowledgement:Earlier I only had a partial proof (I could prove only  LHS$>15/27$) ,But tbanks to River-Li who proposed a stronger Claim(see comments)  and made some corrections the proof is done!
