<s>[INST] 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
$$\frac{8(x+y+z)(xy+yz+xz)}{(x+y)(y+z)(z+x)}.$$
 [/INST] 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 $F=\frac 98$
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