Does anyone have solutions??

Hellooo?!!

The condition is equivalent to $(a+1)(b+1)(c+1)(d+1)=(a-1)(b-1)(c-1)(d-1)$. Let $w=\frac{a+1}{a-1}, x=\frac{b+1}{b-1}, y=\frac{c+1}{c-1}, z=\frac{d+1}{d-1}$. Then the condition is $wxyz=1$. We will prove that they are all positive. Assume for the sake of contradiction w.l.o.g $w \le 0$. Then $a+1$ and $a-1$ can't have the same sign so $a+1 \ge 0>a-1$ which implies $1>a \ge -1$ contradicting $|a|>1$. Thus, $w,x,y,z>0$. Now, $\frac{1}{a-1}=\frac{1}{2} \cdot \frac{2}{a-1}=\frac{1}{2} \cdot (\frac{a+1}{a-1}-1)=\frac{w-1}{2}$. Thus, we shall prove $\frac{w-1}{2}+\frac{x-1}{2}+\frac{y-1}{2}+\frac{z-1}{2}>0$, equivalent to $w+x+y+z>4$ which is trivial by AM-GM.

Thanks for your solution

What is AM-GM

The Inequality of Arithmetic and Geometric Mean