<s>[INST] a inequality with condition For $a,b,c \ge 0$ and $a+b+c \ge m$, I have to prove that $ 6(a^3+b^3+c^3)+9abc \ge m^3$. By power mean inequality
$ 6(a^3+b^3+c^3) \ge \frac {2(a+b+c)^3}{3} \ge \frac{2m^3}{3}$
Then I stuck
 [/INST] By Schur's inequality (look at https://en.wikipedia.org/wiki/Schur%27s_inequality for its statement and a standard proof), 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
$$a^3+b^3+c^3+3abc\geq a(b^2+c^2)+b(c^2+a^2)+c(a^2+b^2)\Rightarrow\\ 2(a^3+b^3+c^3)+3abc\geq (a+b+c)(a^2+b^2+c^2)\geq \frac{(a+b+c)^3}{3}\Rightarrow\\ 6(a^3+b^3+c^3)+9abc\geq m^3.$$
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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