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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!

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

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

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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.

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