Prove $2\sum\limits_{cyc}\,a^{\,3}+ 3\,abc\geqq 3\sum\limits_{cyc}\,a^{\,2}b$ [closed]

For $$a,\,b,\,c\geqq 0$$ and $$b\equiv {\rm mid}\,\{\,a,\,b,\,c\,\}$$. Prove $$2\sum\limits_{cyc}\,a^{\,3}+ 3\,abc\geqq 3\sum\limits_{cyc}\,a^{\,2}b$$ Inspried from $$\lceil$$ Prove $\sum_{cyc}a^3- \sum_{cyc}a^2b \geqq 0$ with $k= 0$ $$\rfloor$$. Good luck you guys! Thanks a real real lot!

closed as off-topic by Martin R, Theo Bendit, mrtaurho, José Carlos Santos, Adrian KeisterMay 23 at 1:23

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Martin R, Theo Bendit, mrtaurho, José Carlos Santos, Adrian Keister
If this question can be reworded to fit the rules in the help center, please edit the question.

It's true for all positives $$a$$, $$b$$ and $$c$$.
Indeed, by Schur $$\sum_{cyc}(2a^2-3a^2b+abc)=\sum_{cyc}(a^3-2a^2b+ab^2)+\sum_{cyc}(a^3-a^2b-a^2c+abc)=$$ $$=\sum_{cyc}a(a-b)^2+\sum_{cyc}a(a-b)(a-c)\geq0.$$
By Schur's inequality, the basic inequality $$3(a^2+b^2+c^2)\geq (a+b+c)^2$$ (which can be proved by Cauchy-Schwarz if we write $$3=1^2+1^2+1^2$$) and the non-negativity of $$a,b$$ and $$c$$ we have that
$$a^3+b^3+c^3+3abc\geq a(b^2+c^2)+b(c^2+a^2)+c(a^2+b^2)\Rightarrow\\ 2(a^3+b^3+c^3)+3abc\geq (a+b+c)(a^2+b^2+c^2)\geq \frac{(a+b+c)^3}{3}.$$
The first line is Schur's inequality. At the second line we just added $$a^3+b^3+c^3$$ at both sides and did the algebra.