very not sure about my "proof" but thought i would post anyway $\displaystyle A(x-y)(x-z) + B(y-z)(y-x) + C(z-x)(z-y) \geq 0.$ (*) Suppose (*). Without loss of generality, $x \geq y \geq z$. [We can repeat the same argument if $x \geq z \geq y$, etc.] Let $a = x-y$ and $b = y-z$ and understand that $a,b$ are positive reals. Then: $(*) \Leftrightarrow A(x-y)(x-y + y-z) + C(x-y + y-z)(y-z) \geq B(x-y)(y-z)$ $\Leftrightarrow Aa(a+b) + C(a+b)b \geq Bab$ $\Leftrightarrow A(\frac{a}{b})^2 + (A+C-B)(\frac{a}{b}) + C \geq 0$ (If $b = 0$, the condition would be just $A \geq 0$). Notice that $\frac{a}{b} \geq 0$. If $A \leq 0$, for extreme values of $a/b$, we have it being negative. So certainly $A > 0$. If there is a positive root, then some segment of the graph of the quadratic is in the 4th quadrant. So we want $- (A+C-B) + \sqrt{A^2 + B^2 + C^2 - 2AB - 2BC - 2CA} < 0$, or $A+C-B > \sqrt{(A+C-B)^2 - 4AC}$, which is true if $AC > 0$. From here, using AMGM it is clear that the condition is $4\sqrt{AC} \geq B$. Because our steps are reversible, it is clear it is sufficent.

I think your proof is wrong! There you can't suppose x>y>z cause it surely loses generality. Remember the inequality is a family of inequalities about A,B,C ,you can't just change the condition of the inequities. My proof is to suppose z=0,the use discriminant of a quadratic equation to get that the square root of A,B,C forms a triangle.

I think that you can suppose $x>y>z$. If you do this, depending on if it's true or not, you can just change $A$, $B$, and $C$ to fit the inequality. Also, It would be nice if you could elaborate on how $\Leftrightarrow Aa(a+b) + C(a+b)b \geq Bab$ $\Leftrightarrow A(\frac{a}{b})^2 + (A+C-B)(\frac{a}{b}) + C \geq 0$

programjames1 wrote: I think that you can suppose $x>y>z$. If you do this, depending on if it's true or not, you can just change $A$, $B$, and $C$ to fit the inequality. Also, It would be nice if you could elaborate on how $\Leftrightarrow Aa(a+b) + C(a+b)b \geq Bab$ $\Leftrightarrow A(\frac{a}{b})^2 + (A+C-B)(\frac{a}{b}) + C \geq 0$ But if we don't suppose x>y>z,from this inequality we can get A^2+B^2+C^2-2AB-2AC-2BC<=0,and A,B,C >=0, it's much stronger than what Singular get.