Solution: Let $f(a,b,c,d)=\dfrac{a-b}{a+b}+\dfrac{b-c}{b+c}+\dfrac{c-d}{c+d}$; $A:\{(x,y,z,u)|1\le x,y,z,u\le 10,f(x,y,z,u)>0\}$, $B:\{(x,y,z,u)|1\le x,y,z,u\le 10,f(x,y,z,u)<0\}$, $C:\{(x,y,z,u)|1\le x,y,z,u\le 10,f(x,y,z,u)=0\}$. Obviously, $card(A)+ card(B)+ card(C)=10^4$ . Now we prove $card(A)= card(B)$. For every $(x,y,z,u) \in A$, $(x,y,z,u) \in A \Leftrightarrow f(x,y,z,u)>0 \Leftrightarrow \dfrac{x-y}{x+y}+\dfrac{y-z}{y+z}+\dfrac{z-u}{z+u}+\dfrac{u-x}{u+x}>0$ $\Leftrightarrow \dfrac{x-u}{x+u}+\dfrac{u-z}{u+z}+\dfrac{z-y}{z+y}+\dfrac{y-x}{y+x}<0 \Leftrightarrow f(x,y,z,u)<0 \Leftrightarrow (x,u,z,y) \in B$ . ∴ $card(A)= card(B)$. Next we find the value of $card(C)$. $(x,y,z,u) \in C \Leftrightarrow \dfrac{xz-yu}{(x+y)(z+u)}= \dfrac{xz-yu}{(y+z)(u+x)} \Leftrightarrow (z-x)(u-y)(xz-yu)=0$ Let $C_1=\{(x,y,z,u)|x=z,1\le x,y,z,u\le 10\}$; $C_2=\{(x,y,z,u)|x\not= z, y=u, 1\le x,y,z,u\le 10\}$; $C_3=\{(x,y,z,u)|x\not= z, y\not= u,xz=yu, 1\le x,y,z,u\le 10\}$ . ∵ there are $9$ disordered quadruples $(a,b,c,d)$ of integers, such that $a\times b=c\times d$ where $1\le a,b,c,d\le 10$ . $1\times 6=2\times 3, 1\times 8=2\times 4, 1\times 10=2\times 5, 2\times 6=3\times 4, 2\times 9=3\times 6,$ $2\times 10=4\times 5, 3\times 8=4\times 6, 3\times 10=5\times 6, 4\times10=5\times 8$. There are $90$ quadruples $(x,y,z,u)$ satisfying $x=y, z=u$ and $x\not= z$. There are $90$ quadruples $(x,y,z,u)$ satisfying $x=u, y=z$ and $x\not= z$. ∴ $card(C_3)=4\times 2\times 9+90+90=252, card(C_1)=1000, card(C_2)=900$. ∴ $card(C)=2152, card(A)=3924$ .

why you used f(x, y, z, u) equals x-y/x+y + y-z/y+z + z-u/z+u + u-x/u+x instead of only x-y/x+y + y-z/y+z + z-u/z+u

jred wrote: Determine the number of ordered quadruples $(x, y, z, u)$ of integers, such that \[\dfrac{x-y}{x+y}+\dfrac{y-z}{y+z}+\dfrac{z-u}{z+u}>0 \textrm{ and } 1\le x,y,z,u\le 10.\] It should be Determine the number of ordered quadruples $(x, y, z, u)$ of integers, such that \[\dfrac{x-y}{x+y}+\dfrac{y-z}{y+z}+\dfrac{z-u}{z+u}+\dfrac{u-x}{u+x}>0 \textrm{ and } 1\le x,y,z,u\le 10.\]

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