Beautiful one.

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Note that $(1-2a)(1-2b)(1-2c) > 0 \implies 4\sum ab > 8abc + 1$. So, $f(a,b,c) = \frac{4\sum ab - 1}{abc} > 8$, and equality cannot be achieved. We will prove $f(a,b,c) \leq 9$. It suffices to show $$\frac{4\sum ab - (\sum a)^3}{abc} \leq 9 \iff \sum a^3 + 3\sum_{\text{sym}} a^2b + 15abc \geq 4(\sum ab)(\sum a)$$$$\iff \sum a^3 + 3abc \geq \sum_{\text{sym}} a^2b$$But the last inequality is Schur's inequality. Thus equality is achieved if and only if $a=b=c=\frac{1}{3}$.

Define $\overline{M}= \{ (a,b,c \in \mathbb{R} : 0 \le a,b,c \le \frac{1}{2} : a+b+c=1 \}$. Note that $\overline{M}$ is bounded and closed, meaning it is a compact set. Hence $f: \overline{M} \rightarrow \mathbb{R}$ has minimum/maximum value in constraint $a+b+c=1$. Let's say $f$ gets its minimum/maximum value at the boundary. Therefore we have either $a=0$ or $a=\frac{1}{2}$. Since $\frac{1}{0}$ is undefined, we can still calculate the value by limits.It is not hard to see that we have to check $a=b=0,c=1$,$a=0,b=c=\frac{1}{2}$ case. $c=1$ case is clearly impossible. therefore for second case which is $a \rightarrow 0, b,c \rightarrow \frac{1}{2}$ we can see $f \rightarrow 8$ If $a \rightarrow \frac{1}{2}$ we have to check $b \rightarrow \frac{1}{2},c \rightarrow 0$ and $b,c\rightarrow \frac{1}{4}$. The first case is same as above.The second case gives $f \rightarrow 8$. Therefore we have finished for this case. The second case is when $a,b,c$ lies on open subset $M$ only, we can bash this with lagrange multipliers since all the functions are differentiable in set $M$. The equation $f_a=f_b=f_c$ gives obvious answer $a=b=c=\frac{1}{3}$. By Graphing the function estimately, we can see there is no such $a\neq b \neq c$ satisfying $f_a=f_b=f_c$. For case $a=b$ we can let $c=1-2b$ and analyze new 1 variable function, which again gives $a=b=c$. and $f(\frac{1}{3},\frac{1}{3},\frac{1}{3})=9$ Therefore $8 < f \le 9$

See also https://artofproblemsolving.com/community/c6h1886311p12854498

Very fun! , we'll prove that $f(M) = (8,9]$. To show this we'll need to do three things, proving the upper and lower bound while also proving that every value is attainable. Upper Bound. $f(a,b,c) \le 9$ for all $(a,b,c) \in M$ Proof. First, let's use $(a+b+c)^2=1$ to transform $$f(a,b,c)=\frac{4(ab+bc+ca)-1}{abc}=\frac{1-2(a^2+b^2+c^2)}{abc}$$. What we want to prove is $$2(a^2+b^2+c^2)+9abc\ge 1$$, using $a+b+c=1$ to homogenize the inequality yields $$2(a^2+b^2+c^2)(a+b+c)+9abc\ge (a+b+c)^3$$$$\Rightarrow 2\sum_{cyc}a^3+2\sum_{sym}a^2b+9abc \ge \sum_{cyc}a^3 + 3\sum_{sym}a^2b +6ab$$$$\Rightarrow \sum_{cyc}a^3+3abc\ge \sum_{sym}a^2b$$. Which is true by Schur's Inequality. As desired. Lower Bound. $f(a,b,c) > 8$ for all $(a,b,c) \in M$ Proof. Note that $$f(a,b,c)=\frac{4(ab+bc+ca)-1}{abc}$$What we want to prove is $$1+8abc<4(ab+bc+ca)$$. Homogenizing the inequality and cancelling yields (same as before) $$(a+b+c)^3+8abc<4(ab+bc+ca)(a+b+c)$$$$\Rightarrow \sum_{cyc}a^3+ 3\sum_{sym}a^2b+14abc<4\sum_{cyc}a^2b+12abc$$$$\Rightarrow \sum_{cyc}a^3+2abc < \sum_{sym}a^2b$$. Note that $a,b,c$ are the sides of a triangle ($b+c>a$, and similarly for others), let $a=x+y,b=y+z,c=z+x$ where $x,y,z>0$. We have, $$ \sum_{cyc}a^3+2abc=\sum_{cyc} x^3+3x^2y+3xy^2+y^3 + 2\sum_{sym}x^2y+4xyz $$$$\Rightarrow \sum_{cyc}a^3+2abc=2\sum_{cyc}x^3 +5\sum_{sym}x^2y+4xyz$$. While $$\sum_{sym}a^2b = \sum_{sym}y^3+3x^2y+y^2z+zx^2+2xyz$$$$\Rightarrow \sum_{sym}a^2b = 2\sum_{cyc}x^3+5\sum_{sym}x^2y+12xyz$$. So, $$\Rightarrow \sum_{cyc}a^3+2abc < \sum_{sym}a^2b$$. As desired. Attainability. Every value in $(8,9]$ is attainable Proof. Pick $a=b=c=\frac{1}{3}$ to get $f(a,b,c)=9$. Next, pick $a=b=x$ and $c=1-2x$ where $x \in \left( \frac{1}{3},\frac{1}{2} \right)$, it obeys the restriction $0<a,b,c <\frac{1}{2}$. Then, $$f(a,b,c)=\frac{4(ab+bc+ca)-1}{abc}=\frac{-12x^2+8x-1}{x^2(1-2x)}=\frac{6x-1}{x^2}$$. Define $g(x)=\frac{6x-1}{x^2}$ where $x \in \left( \frac{1}{3},\frac{1}{2} \right)$ , then $$g'(x)=-6x^{-2}+2x^{-3}=2x^{-2}(x^{-1}-3)<0$$. So $g$ is strictly decreasing and clearly continuous at $\left( \frac{1}{3},\frac{1}{2} \right)$. We have $g\left( \frac{1}{3}\right)=9$ and $g\left( \frac{1}{2} \right)=8$. Hence every value in $(8,9)$ is attainable by $g$, and our job is finished . Hence $\alpha=8$ and $\beta=9$ where $\alpha$ is an infimum while $\beta$ is a maximum.

First,let's use $a+b+c=1$ to transform $$f(a,b,c)=6+\sum_{cyc}\frac{a^2+b^2-c^2}{ab}$$Note that $a,b,c$ are the lengths of the sides of an arbitrary triangle, so $cos C=\frac{a^2+b^2-c^2}{2ab}$, therefore $$f(a,b,c)=6+\sum_{cyc}cos A$$and in a triangle $1 < \sum_{cyc}cos A \le \frac{3}{2}$ then $8 < f(a,b,c) \le 9$

amogususususus wrote: Very fun! , we'll prove that $f(M) = (8,9]$. To show this we'll need to do three things, proving the upper and lower bound while also proving that every value is attainable. Upper Bound. $f(a,b,c) \le 9$ for all $(a,b,c) \in M$ Proof. First, let's use $(a+b+c)^2=1$ to transform $$f(a,b,c)=\frac{4(ab+bc+ca)-1}{abc}=\frac{1-2(a^2+b^2+c^2}{abc}$$. What we want to prove is $$2(a^2+b^2+c^2)+9abc\ge 1$$, using $a+b+c=1$ to homogenize the inequality yields $$2(a^2+b^2+c^2)(a+b+c)+9abc\ge (a+b+c)^3$$$$\Rightarrow 2\sum_{cyc}a^3+2\sum_{sym}a^2b+9abc \ge \sum_{cyc}a^3 + 3\sum_{sym}a^2b +6ab$$$$\Rightarrow \sum_{cyc}a^3+3abc\ge \sum_{sym}a^2b$$. Which is true by Schur's Inequality. As desired. Lower Bound. $f(a,b,c) > 8$ for all $(a,b,c) \in M$ Proof. Note that $$f(a,b,c)=\frac{4(ab+bc+ca)-1}{abc}$$What we want to prove is $$1+8abc<4(ab+bc+ca)$$. Homogenizing the inequality and cancelling yields (same as before) $$(a+b+c)^3+8abc<4(ab+bc+ca)(a+b+c)$$$$\Rightarrow \sum_{cyc}a^3+ 3\sum_{sym}a^2b+14abc<4\sum_{cyc}a^2b+12abc$$$$\Rightarrow \sum_{cyc}a^3+2abc < \sum_{sym}a^2b$$. Note that $a,b,c$ are the sides of a triangle ($b+c>a$, and similarly for others), let $a=x+y,b=y+z,c=z+x$ where $x,y,z>0$. We have, $$ \sum_{cyc}a^3+2abc=\sum_{cyc} x^3+3x^2y+3xy^2+y^3 + 2\sum_{sym}x^2y+4xyz $$$$\Rightarrow \sum_{cyc}a^3+2abc=2\sum_{cyc}x^3 +5\sum_{sym}x^2y+4xyz$$. While $$\sum_{sym}a^2b = \sum_{sym}y^3+3x^2y+y^2z+zx^2+2xyz$$$$\Rightarrow \sum_{sym}a^2b = 2\sum_{cyc}x^3+5\sum_{sym}x^2y+12xyz$$. So, $$\Rightarrow \sum_{cyc}a^3+2abc < \sum_{sym}a^2b$$. As desired. Attainability. Every value in $(8,9]$ is attainable Proof. Pick $a=b=c=\frac{1}{3}$ to get $f(a,b,c)=9$. Next, pick $a=b=x$ and $c=1-2x$ where $x \in \left( \frac{1}{3},\frac{1}{2} \right)$, it obeys the restriction $0<a,b,c <\frac{1}{2}$. Then, $$f(a,b,c)=\frac{4(ab+bc+ca)-1}{abc}=\frac{-12x^2+8x-1}{x^2(1-2x)}=\frac{6x-1}{x^2}$$. Define $g(x)=\frac{6x-1}{x^2}$ where $x \in \left( \frac{1}{3},\frac{1}{2} \right)$ , then $$g'(x)=-6x^{-2}+2x^{-3}=2x^{-2}(x^{-1}-3)<0$$. So $g$ is strictly decreasing and clearly continuous at $\left( \frac{1}{3},\frac{1}{2} \right)$. We have $g\left( \frac{1}{3}\right)=9$ and $g\left( \frac{1}{2} \right)=8$. Hence every value in $(8,9)$ is attainable by $g$, and our job is finished . Hence $\alpha=8$ and $\beta=9$ where $\alpha$ is an infimum while $\beta$ is a maximum. can you give a formal proof of function $g$'s continuity? i think you should prove that the limit exists