Simple generalization, right?

Take $y=2$ and $z=1$ (noting you cannot then take $x=-2$ or $x=-1$ ). I get the equation $\frac{x}{3}+\frac{2}{x+1}+\frac{1}{x+2}=k$ . I cleared the denominators, and combined like terms, which resulted in the equation: $x^3+3(1-k)x^2+(11-9k)x+15-6k$ . Since the equation is a polynomial of odd degree in $x$ (cubic), there exists at least one real value of $x$ that is a solution. Also, there does not exist a $k$ that gives a root of $x=-2$ , and likewise for $x=-1$ , so we needn't worry about those. This can be seen by substituting them both ( $x=-2,-1$ ) into the cubic, and seeing that $k$ vanishes, and each side of the equation gives a different value, resulting in a contradiction. So, this is enough to conclude that for any real number $k$ , there exists an ordered triple (of the form $(x,y,z)=(x,2,1)$ , for some $x$ ) that solves the equation $\frac{x}{y+z}+\frac{y}{x+z}+\frac{z}{x+y}=k$ .

Lets first suppose that indeed for all real numbers $k$ we can find some real numbers such that $\frac{x}{y+z} + \frac{y}{z+x} + \frac{z}{x+y} = k$ . If we now add 3 to both sides then we would have that $\frac{x}{y+z} + 1 + \frac{y}{z+x} + 1 + \frac{z}{x+y} + 1 = k + 3$ but the left hand side now can be written as $(x+y+z) \left( \frac{1}{y+z} + \frac{1}{z+x} + \frac{1}{x+y} \right) = k + 3$ . Let's now substitute $u = k+3$ because the statement "all k have solutions" is equivalent to the statement "all u have solutions" to get a new equation, $(x+y+z) \left( \frac{1}{y+z} + \frac{1}{z+x} + \frac{1}{x+y} \right) = u$ . Now, I want to find solutions or at least prove solutions exist for all u. I have now simplified the algebra, but it can be simplified further by picking $x,y,z$ such that $x + y + z = 1$ . The candidates I chose were $x=x, y = -x + a, z = 1-a$ where a is some arbitrary real number. If we now substitute this in we get the equation:

$\frac{1}{-x + 1} + \frac{1}{x+1-a} + \frac{1}{a} = u$ . Now, by studying this function and its first derivative one finds three things. First, at the values where the function blows up it runs through all the negative numbers, so solutions for negative u are trivial (notice that at these values the function does not run through all the positive numbers though). Second, there is a root at $x = \frac{1}{2} \left((-3a^2 + 4a + 4)^{\frac{1}{2}} + a \right)$ (for small enough a) and third, that for small enough a (taking 0 < a < 1 is enough), the function increases after that point asymptotically towards the limit $\lim_{x \to \infty} \frac{1}{-x + 1} + \frac{1}{x+1+a} + \frac{1}{a} = \frac{1}{a}$ so by picking a small enough $a$ we can find a function that will yield a solution for all values in the interval $[0, \frac{1}{a} )$ so effectively for any given u, one can pick a small enough to find a function that will yield a solution.

For example, to find a solution to the original problem when $k=0$ then put $u = 3$ (because k=u-3) and choose $a=0.05$ . Then a solution is approximately $x=1.05715, y= -1.00715, z=0.95$ . But because one may pick smaller $a$ and find new solutions what has actually been proven is a slightly stronger statement, that all k have infinitely many solutions.

Let $x = a$ and $y = b$ , where $a,b \in \mathbb{R}$ and $z \neq -a, z \neq -b, a + b \neq 0, (a \neq 0 \land b \neq 0 )$ and $a \neq b \implies \dfrac{a}{z + b} + \dfrac{b}{z + a} + \dfrac{z}{a + b} = k$

$\implies z^3 + (a + b)(1 - k) z^2 + ((a + b)^2 (1 - k) + ab) z + (a + b)(a^2 + b^2 + abk) = 0.$

Let $F(z) = z^3 + (a + b)(1 - k) z^2 + ((a + b)^2 (1 - k) + ab) z + (a + b)(a^2 + b^2 + abk) \implies F(z)$ has at least one real root( since a cubic polynomial with real coefficients has at least one real root ).

Choose the real root $z = g(a,b,k)$ , then for each $k \in \mathbb{R}, (x,y,z) = (a,b,g(a,b,k))$ satisfies $\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y}=k.$ .

Taking into account that the left-hand side is defined in the region $D=\{(x,y,z)\mid x\ne 0,y\ne 0,\,z\ne0 , y\ne-z,z\ne-x,x\ne-y\}$ the given equation reduces to the algebraic equation of third degree in $x$ , i.e. $x^3 + x^2 (y + z) (1-k) + x( (y^2+ z^2)(1 - k) + y z(3 - 2 k) ) + y^2(y + z - k z) + z^2(y + z - k y) = 0,$ which always has one solution, for any real values of $k$ , and values of $y, z$ such that $y\ne 0, z\ne0,y+z\ne 0$ .

I believe it is like a partial fraction and so if you perform the operation you get real nos and since real covers a lot the answers are covered