Given three side-lengths $a, b, c$ of a triangle. Prove that $$a^{2}b\left ( a- b \right )+ b^{2}c\left ( b- c \right )+ c^{2}a\left ( c- a \right )\geq 3\left ( a+ b- c \right )c\left ( a- b \right )\left ( b- c \right )$$ Source: StackMath/@haidangel ft.@tthnew
I used discriminant to create this inequality, also $constant\!:\!=\!\!3$ is the best here. See_ on.StackMath, that's what I'm doing research on, of course this inequality is a result. Not an answer. I think we use Bernhard Leeb's result to help a real lot $$a^{2}b\left ( a- b \right )+ b^{2}c\left ( b- c \right )+ c^{2}a\left ( c- a \right ):= \left ( c+ a- b \right )\left ( c- a \right )^{2}b- \left ( b+ c- a \right )\left ( a- b \right )\left ( b- c \right )c$$
Just a comment but it's too long for a comment.
Some try to proof: Since $a,b,c$ be three sid-lengths of a triangle, if $(a-b)(b-c)<0$ it's obvious.
So, we only need to consider this inequality when $(a-b)(b-c)\geq 0\to b\equiv \text{mid}\{a,b,c\}.$
But I can't prove this.
By the way if we change $3\to \frac{12}{5}=2.4,$ it's much easy to prove (for the weaker version).
Indeed, let $a=x+y,b=y+z,c=z+x$ and multiply our inequality with $x+y+z.$ The inequality becomes: $$\frac{1}{20} \left( x-y \right)^2z^2 \left( 15x+31y \right) +\dfrac{1}{20} \left( 2x-y-z \right)^2y \left( 20x^2+19xy+46zx+37 z^2 \right) $$ $$+{\frac {21}{20}} \left( zx+y^2-2yz \right)^2x+\frac {17}{10} \left( x-z \right)^2y^2z+\frac {3}{20} \left( z+x-2y \right) ^2z ^2y$$ $$+\frac{2}{5} \left( x^2-yz \right)^2 \left( 2y+5 z \right) +\frac{1}{5} \left( y-z \right) ^{2}x^3\ge 0.$$
According to Mr. RiverLi's idea, After let $a=x+y,b=y+z,c=z+x$ then $f(a,b,c)=f(x+y,y+z,z+x)\equiv f(x),$ and we only need to prove $$\Delta_x=-16 y^2 (y-z)^4 (y+z) (3 y+z) \left(y^2-4 y z+27 z^2\right)\le 0,$$ which is clearly true. I'm still try to find another way.