Given three non-negative $a,b,c$ so that $c$ between $a$ and $b$. Prove that $2\sum\limits_{cyc}a^{3}+3abc\geqq3\sum\limits_{cyc}a^{2}b$ .

Question

Given three non-negative $a, b, c$ so that $c$ between $a$ and $b$. Prove that $2\sum\limits_{cyc}a^{3}+ 3abc\geqq 3\sum\limits_{cyc}a^{2}b$

Inspried from: https://math.stackexchange.com/a/3264953/688846, after using Ravi-subs for $$(\sum\limits_{cyc}a^{3}- \sum\limits_{cyc}a^{2}b)\geqq 0$$ by $a= a+ b, b= b+ c, c= c+ a$, equal to $$(\sum\limits_{cyc}a^{3}+ \sum\limits_{cyc}a^{2}b- 2\,abc)\geqq 0$$ And again, we have $$5(a^{3}+ b^{3}+ c^{3}- 3\,abc)= 3(\sum\limits_{cyc}a^{3}+ \sum\limits_{cyc}a^{2}b- 2\,abc)+ (\underbrace{2\sum\limits_{cyc}a^{3}+ 3abc- 3\sum\limits_{cyc}a^{2}b}_{\geqq 0(that's\,all\,we\,need\,to\,prove\,!)})\geqq 0$$

Answer 1

It's true for all positives $a$, $b$ and $c$.

Indeed, by Schur $$\sum_{cyc}(2a^2-3a^2b+abc)=\sum_{cyc}(a^3-2a^2b+ab^2)+\sum_{cyc}(a^3-a^2b-a^2c+abc)=$$ $$=\sum_{cyc}a(a-b)^2+\sum_{cyc}a(a-b)(a-c)\geq0.$$

Answer 2

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.