I am looking for a proof. Here is my question.
Problem. Given non-negative real numbers $a,b,c$ satisfying $a+b+c=3.$ Prove that$$\color{black}{a^2+b^2+c^2+3\ge 2\sqrt{2(a^3b+b^3c+c^3a)+3abc}. }\tag{1}$$
Source: own
Equality holds iff $(a,b,c)=(1,1,1);(2,1,0)$ or its cyclic permutations.
I came up with this inequality from a weaker one $$\color{black}{a^2+b^2+c^2+3\ge 2\sqrt{2}\cdot\sqrt{a^3b+b^3c+c^3a}.}\tag{2}$$
Here is my two ideas for $(2).$
Idea 1.
Denote $a+b+c=3=p; ab+bc+ca=q; abc=r.$
Verifying $\sum_{cyc}a^3b\longrightarrow \sum_{cyc}a^2b,$
$$a^3b=\sum_{cyc}a\sum_{cyc}a^2b-\sum_{cyc}a^2b^2-abc\sum_{cyc}a=3\sum_{cyc}a^2b-q^2+3r$$
By a well-known result $$\boxed{a^2b+b^2c+c^2a+abc\le \frac{4}{27}(a+b+c)^3, \forall a,b,c\ge 0. }\tag{*}$$
It remains to prove $$\color{black}{12-2q\ge 2\sqrt{2}\sqrt{12-q^2}.}\iff (q-2)^2\ge 0.$$
Remark $1$.
The $(*)$ doesn't help for $(1)$ since $$\color{black}{12-2q\ge 2\sqrt{2(12-q^2)+3r}}$$is not true when $a=b\rightarrow 1^{-}.$
Idea 2.
We can use directly the following result (maybe own) $$\boxed{\color{black}{4(a+b+c)^4\ge 27[a^3b+b^3c+c^3a+(ab+bc+ca)^2], \forall a,b,c\ge 0. }}\tag{4}$$
It follows by BW.
Remark $2$.
The $(4)$ doesn't help for $(1)$ since both $(3)$ and $(4)$ is exactly same!
Equality holds at $a=b=c$ or $a=0;b=2c$ and permutations. Recently, I've realised that $$(a+b+c)(a^2b+b^2c+c^2a+abc)=a^3b+b^3c+c^3a+(ab+bc+ca)^2\le \frac{4}{27}(a+b+c)^4.$$
In conclusion, the inequality $(2)$ is proven but the stronger version $(1)$ is not complete.
I think there's at least two proofs for $(1)$ and it will be posted soon. Now, I'd like to find nice ideas for starting inequality.
Any ideas and comments are welcome. Thank you for your interest.
Remark $3.$ Here is some relevant inequalities.
I check that$$\color{black}{\frac{a^2+b^2+c^2+3+(-6+2\sqrt{6})abc}{8}\ge \sqrt{\frac{a^3b+b^3c+c^3a}{8}}}$$which save two equality cases but it's not true.
Also, the following inequality is true and I hope it will be useful for $(1).$
$$\color{black}{2(a^2+b^2+c^2)\ge a^3b+b^3c+c^3a+abc+2.}$$
Equality also holds iff $(a,b,c)=(1,1,1);(2,1,0)$ or its cyclic permutations.
It is obvious by BW. Hope there's a good proof like d8g3n1v9
's idea, similar problem.
See also Riverli's answer for an idea.
Best Answer
We have: \begin{align*} \left[a^2+b^2+c^2+\frac{1}{3}\left(a+b+c \right)^2 \right]^2 -8\left(a^3b+b^3c+c^3a \right)-4abc\left(a+b+c \right) \\ = \frac{2}{27} \sum_{\text{cyc}}\left( a-2b+4c\right)^2 \left(b+c-2a \right)^2, \end{align*} which directly implies the result.