Extreme values of $\frac{(a+b+c)(2ab+2bc+2ca-a^2-b^2-c^2)}{abc}$

Question

Let $a,b,c$ be side lengths of a triangle. What are the minimum and maximum of $$\frac{(a+b+c)(2ab+2bc+2ca-a^2-b^2-c^2)}{abc}?$$

When $a=b=c$, the value is $9$. In addition, we can write $a=x+y,b=y+z,c=z+x$ since they are side lengths of a triangle. The expression becomes $$\frac{8(x+y+z)(xy+yz+xz)}{(x+y)(y+z)(z+x)}.$$

Answer 1

We will prove that $$8 \leqslant \frac{8(x+y+z)(xy+yz+zx)}{(x+y)(y+z)(z+x)} \leqslant 9$$

For the Left Hand Side:

$$(x+y+z)(xy+yz+zx) - (x+y)(y+z)(z+x)=xyz \geqslant 0$$

For the Right Hand Side

\begin{align*} \ & 9(x+y)(y+z)(z+x)-8(x+y+z)(xy+yz+zx) \\ &=x^2y+x^2z+xy^2+xz^2+y^2z+yz^2-6xyz \\ &\geqslant 0 \end{align*} which is true according to AM-GM.

So the minimum value is $8$ , achieve when $x=0$ and cyclic permutation
The maximum value is $9$, achieve at $x=y=z$

Answer 2

For the maximum values, we could also show it using derivatives.

Consider the function $$F=\frac{(x+y+z)(xy+yz+zx)}{(x+y)(y+z)(z+x)}$$ Compute derivatives; after simplifications, we get $$F'_x=\frac{y z \left(y z-x^2\right)}{(x+y)^2 (x+z)^2 (y+z)}$$ $$F'_y=\frac{x z \left(x z-y^2\right)}{(x+y)^2 (x+z) (y+z)^2}$$ $$F'_z=\frac{x y \left(x y-z^2\right)}{(x+y) (x+z)^2 (y+z)^2}$$ If none of $x,y,z$ is zero, then, for cancelling each derivative, we have $$yz-x^2=xz-y^2=xy-z^2=0$$ which admits as only real solutions $x=y=z=1$ for which $F=\frac 98$