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
$$\frac{8(x+y+z)(xy+yz+xz)}{(x+y)(y+z)(z+x)}.$$

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, $$\max\frac{(a+b+c)(2ab+2bc+2ca-a^2-b^2-c^2)}{abc}=9.$$
