Extreme values of $\frac{(a+b+c)(2ab+2bc+2ca-a^2-b^2-c^2)}{abc}$ 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
$$\frac{8(x+y+z)(xy+yz+xz)}{(x+y)(y+z)(z+x)}.$$

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 $x=y=z$
