Wrong

Complete_quadrilateral wrote: WLOG $a\ge b,c$ Case 1: $b \ge c$ $c\le a, \frac{4}{a}\le \frac{4}{b}$ Huh? Note that $a,b,c$ are not necessarily positive...

Tintarn wrote: Complete_quadrilateral wrote: WLOG $a\ge b,c$ Case 1: $b \ge c$ $c\le a, \frac{4}{a}\le \frac{4}{b}$ Huh? Note that $a,b,c$ are not necessarily positive... wlog doesnt give us result

(2,-1,-4) = (a,b,c) is a sols btw

My bad I thought they were positive

Suppose that $a=b$. Then $\frac{4}{b}=\frac{4}{c}$ and so $a=b=c$ which is a solution. And so $a \neq b \neq c \neq a$. From the first equation we get $a-b = \frac{4(b-c)}{bc}$ and similarly $b-c = \frac{4(c-a)}{ca}, c-a = \frac{4(a-b)}{ab}$. Multiplying these three we get $(abc)^2=2^6$ or $|abc| = 8$.Without loss of generality, we can assume $abc=8$ because if $(a,b,c)$ is a solution, $(-a,-b,-c)$ is also a solution. Let $a+\frac{4}{b} =x$. Multiplying we get $x^3=12x+16$ which yields $x=-2,x=4$. the rest is easy.

Parsia-- wrote: Suppose that $a=b$. Then $\frac{4}{b}=\frac{4}{c}$ and so $a=b=c$ which is a solution. And so $a \neq b \neq c \neq a$. From the first equation we get $a-b = \frac{4(b-c)}{bc}$ and similarly $b-c = \frac{4(c-a)}{ca}, c-a = \frac{4(a-b)}{ab}$. Multiplying these three we get $(abc)^2=2^6$ or $|abc| = 8$.Without loss of generality, we can assume $abc=8$ because if $(a,b,c)$ is a solution, $(-a,-b,-c)$ is also a solution. Let $a+\frac{4}{b} =x$. Multiplying we get $x^3=12x+16$ which yields $x=-2,x=4$. the rest is easy. how did you get from $a + \frac{4}{b} = x$ to $x^3 = 12x + 16$?

I claim that the answers are the permutations of the following: $\boxed{\left(\pm2, \mp4, \mp1\right)}$, $\boxed{\left(x, x, x\right)}$, where $x\in\mathbb{R}-\{0\}$, and $\boxed{\left(\frac{-2(x+2)}{x}, x, \frac{-4}{x+2}\right)}$, $\boxed{\left(\frac{2(x-2)}{x}, x, \frac{-4}{x-2}\right)}$, where $x\in\mathbb{R}-\{0\}\land x^2-4\neq 0$. Note that $a, b, c\in\mathbb{R}-\{0\}$. If any two of $a, b$ or $c$ are equal, then $a=b=c$. Suppose $a=b$. This suggests that $\frac{4}{b}=\frac{4}{c}$ or $b=c$. Similarly, the result follows for the rest of the pairs as well. So, $\boxed{\left(x, x, x\right)}$, where $x\neq 0$ is a trivial solution to the equation. Henceforth, suppose none of $a$, $b$ or $c$ are equal. Rewrite the equations as $$a-b=\frac{4(b-c)}{bc}$$$$b-c=\frac{4(c-a)}{ca}$$$$c-a=\frac{4(a-b)}{ab}$$Taking their product, we have $\frac{64}{(abc)^2}=1$ or $abc=\pm8$. Case 1: $abc=8$. If $b^2-4=0$, then we get the solution $\boxed{\left(-4, 2, -1\right)}$ and its permutations. Otherwise, $a+\frac{4}{b}=b+\frac{4}{c} \implies abc+4c=b^2c+4b \implies c=\frac{8-4b}{b^2-4}=\frac{-4}{b+2}$ and $a=\frac{8}{bc}=\frac{-2(b+2)}{b}$. Therefore, $\boxed{\left(\frac{-2(x+2)}{x}, x, \frac{-4}{x+2}\right)}$, where $x^2-4\neq 0$, and its permutations are solutions to the equation. $\blacksquare$ Case 2: $abc=-8$. If $b^2-4=0$, then we get the solution $\boxed{\left(4, -2, 1\right)}$ and its permutations. Otherwise, in a similar fashion we get $c=\frac{-4}{b-2}$ and $a=\frac{2(b-2)}{b}$. Therefore, $\boxed{\left(\frac{2(x-2)}{x}, x, \frac{-4}{x-2}\right)}$, where $x^2-4\neq 0$, and its permutations are solutions to the equation. $\blacksquare$ $\square$