About a cyclic inequality $a^2+b^2+c^2+3\ge 2\sqrt{2(a^3b+b^3c+c^3a)+3abc}.$

Question

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

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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.

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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.

Answer 1

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.

Answer 2

My proof using BW for $(1).$

After homogenization$$\color{black}{3(a^2+b^2+c^2)+(a+b+c)^2\ge 6\cdot\sqrt{2(a^3b+b^3c+c^3a)+abc(a+b+c)}.}$$ WLOG, assuming $a=\min\{a,b,c\}.$ Let $b=a+u; c=a+v (a,u,v\ge 0).$

We obtain$$\left[3(a^2+b^2+c^2)+(a+b+c)^2\right]^2-36\left[2(a^3b+b^3c+c^3a)+abc(a+b+c)\right]$$ $$=4\cdot\left[9a^2(u^2-uv+v^2)+3a(2u^3-9u^2v+9uv^2+2v^3)+(u-2v)^2(4u^2+2uv+v^2)\right].$$A big trouble is here$$2u^3-9u^2v+9uv^2+2v^3=2v^3+u(u-3v)(2u-3v).$$We can continue the approach due to discriminant theory.

Indeed, let $u=xv; x\ge 0.$ Now, we consider $$f(a)=9a^2(u^2-uv+v^2)+3a(2u^3-9u^2v+9uv^2+2v^3)+(u-2v)^2(4u^2+2uv+v^2)$$is a quadratic equation of $a$ where $$\Delta_{a}=9\left[(2u^3-9u^2v+9uv^2+2v^3)^2-4(u-2v)^2(4u^2+2uv+v^2)(u^2-uv+v^2)\right]$$ $$=-27v^6 (x - 2)^2 (x + 1)^2 (2 x - 1)^2\le 0.$$Thus $f(a)\ge 0$ and we are done!