Given three numbers $a, b, c$ so that $a+ b, b+ c, c+ a\geqq 0$. Prove that $k= 0$ is the only $constant$ so that
$$(\sum\limits_{cyc}a^{3}- \sum\limits_{cyc}a^{2}b)- k(a- b)(a- c)(b+ c)\geqq 0$$
This inequality could be generalized from the problem as follow for same conditions with that above
$$(\sum\limits_{cyc}a^{3}- \sum\limits_{cyc}a^{2}b)\geqq 0$$
$$\because(\sum\limits_{cyc}a^{3}- \sum\limits_{cyc}a^{2}b)= (a- b)(a- c)(a+ c)+ (b+ c)(b- c)^{2}\geqq 0$$
I've found only $0$ by discriminant, I tried to subs $a, b, c$ as same as an old problem but unsuccessfully
Best Answer
Let $a+b=z$, $a+c=y$ and $b+c=x$.
Thus, $x$, $y$ and $z$ are non-negatives and we obtain: $$\sum_{cyc}(x^2y-xyz)\geq kx(x-y)(x-z),$$ which for $x=1$ and $y=z=0$ gives $k\leq0.$
Now, let $z=0$ and $y>x$.
Thus, $$y\geq k(x-y)$$ or $$k\geq\frac{y}{x-y},$$ which gives $$k\geq\max_{y>x\geq0}\frac{y}{x-y}=-1.$$
Now, show that $k=-1$ is valid and show also that for all $-1\leq k\leq0$ your inequality is true.