The motivation of this question is to look if there is such solution in rational number to the identity which mentioned here, I have done many attempts using Wolfram Alpha to find such pairs of rationals $(x,y,z)$ for which $\frac{x}{y+z}+\frac{y}{x+z}+\frac{z}{y+x}=1$ but I failed even I believed that there are no such solutions?

$\begingroup$ Relevant question: mathoverflow.net/questions/227713/… $\endgroup$ – Thomas Browning May 2 '20 at 2:46

6$\begingroup$ The question that Thomas Browning linked has an answer by Michael Stoll pointing out that solutions to this equation lie on an elliptic curve of rank 0, so the only solutions are the trivial ones. $\endgroup$ – S. Carnahan♦ May 2 '20 at 2:54

6$\begingroup$ Does this answer your question? Estimating the size of solutions of a diophantine equation $\endgroup$ – ARG May 2 '20 at 6:11
Let, $(x,y,z)$ be real. Then $a=x+y, b=y+z, c=z+x$ all are reals.
Now, $\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y}=1$
Or, $(x+y+z)(\frac{1}{y+z}+\frac{1}{z+x}+\frac{1}{x+y})=4$
Or,
$((x+y)+(y+z)+(z+x))(\frac{1}{y+z}+\frac{1}{z+x}+\frac{1}{x+y})=8$
Or, $(a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})=8$
Or,
$(\sqrt{\frac{a}{b}}\sqrt{\frac{b}{a}})^2+(\sqrt{\frac{b}{c}}\sqrt{\frac{c}{b}})^2+(\sqrt{\frac{c}{a}}\sqrt{\frac{a}{c}})^2=1$
So, $a,b,c$ shouldn't be all positive or all negative.
Suppose, $c$ is negative then we change the equation as below
$(a+bc)(\frac{1}{a}+\frac{1}{b}\frac{1}{c})=8$ taking $c>0$.
Let, $ac=d>0, a>c>b$. Then the equation becomes $(d+b)(\frac{1}{b}\frac{d}{m}), m=ac$...$(1)$
Or, let $ca=d>0, a>c>b$. Or, $(bd)(\frac{1}{b}+\frac{d}{m})$....$(2)$
All other cases are symmetrically equivalent.
These two cases similarly implies if $\sqrt{(\frac{b}{m}\frac{1}{b})^2\frac{28}{m})}$ is rational, then there are rational solutions. It is only left to show whether there exists non zero rational $b,m$ such that $\sqrt{((b²m)^228b^2m)}=\text{rational}$.
Now, $(b²m)^228b^2m)= {m}^2(t^230t+1)={m}^2(t\alpha)(t\alpha*)$
where, $l, t $rational and $t=\frac{b^2}{m}$. And $\alpha=15+\sqrt{224}$ and $\alpha*=15\sqrt{224}$.
Its our next job to find whether there is such $t$ such that $(t\alpha)(t\alpha*)=r^2$ for some rational $r$....$(3)$
 $t=30, r=1$ is a trivial solution, $b^2=30m=30ac$ , but this doesn't give any solution as $b=\sqrt{30ac} \nless \text{both} a,c$ which is required in $(1)$ or $(2)$.
$(1), (2)$ and $(3)$ requires $t<15\sqrt{224}$.

$\begingroup$ square root of a real number is not necessarily real. $\endgroup$ – GTA May 2 '20 at 4:18