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
$$\frac{8(x+y+z)(xy+yz+xz)}{(x+y)(y+z)(z+x)}.$$
 A: 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 $x=y=z$
A: 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 $F=\frac 98$
A: If $a=b=1$ and $c\rightarrow2^-$ so $$\frac{(a+b+c)(2ab+2bc+2ca-a^2-b^2-c^2)}{abc}\rightarrow8,$$
but $$\frac{(a+b+c)(2ab+2bc+2ca-a^2-b^2-c^2)}{abc}>8$$ it's
$$\sum_{cyc}\left(-a^3+a^2b+a^2c-\frac{2}{3}abc\right)>0$$ or
$$(a+b-c)(a+c-b)(b+c-a)>0,$$
which says that the minimum does not exist and
$$\inf\frac{(a+b+c)(2ab+2bc+2ca-a^2-b^2-c^2)}{abc}=8$$
Let $a=b=c$.
Hence, $$\frac{(a+b+c)(2ab+2bc+2ca-a^2-b^2-c^2)}{abc}=9,$$
but $$\frac{(a+b+c)(2ab+2bc+2ca-a^2-b^2-c^2)}{abc}\leq9$$ it's
$$\sum_{cyc}(a^3-a^2b-a^2c+abc)\geq0,$$
which is Schur.
Thus, $$\max\frac{(a+b+c)(2ab+2bc+2ca-a^2-b^2-c^2)}{abc}=9.$$
