$x\sqrt{1-\frac{y(x+1)}{(x+y+1)^2}} = y\sqrt{1-\frac{z(y+1)}{(y+z+1)^2}} = z\sqrt{1-\frac{x(z+1)}{(z+x+1)^2}}$ 
If $x,y,z > 0$, and $$x\sqrt{1-\frac{y(x+1)}{(x+y+1)^2}} = y\sqrt{1-\frac{z(y+1)}{(y+z+1)^2}} = z\sqrt{1-\frac{x(z+1)}{(z+x+1)^2}}$$
Prove $x=y=z$.

I try it by Mathematica but can not solve, and Maple give a solution, but so complicated. I want to use inequality, but never find the solution.
 A: Let $$f(x,y)=x^2\Bigg(1-\frac{(x+1)y}{(x+y+1)^2}\Bigg)$$
Then the hypothesis says that $f(x,y)=f(y,z)=f(z,x)$. Note that for $x\neq y$ and $x,y$ positive,
$$
\frac{f(y,x)-f(x,y)}{y-x}=\frac{x^3+y^3+2(x^2+y^2)+xy(2(x+y)+3)+x+y}{(x+y+1)^2} > 0 \tag{1}
$$  
Similarly, if we expand $\frac{f(x_2,y)-f(x_1,y)}{x_2-x_1}$, we find that it is equal to $\frac{P(x_1,x_2,y)}{(x_1+y+1)^2(x_2+y+1)^2}$ where $P$ is a polynomial in $x_1,x_2,y$ with positive coefficients. So for $x_1\neq x_2$ and $x_1,x_2,y$ positive, we have
$$
\frac{f(x_2,y)-f(x_1,y)}{x_2-x_1}> 0 \tag{2}
$$  
We can now proceed with the proof. Suppose that $x<y$. Then $f(y,x)>f(x,y)$ by (1),but $f(x,y)=f(z,x)$ by hypothesis, so $f(y,x)>f(z,x)$. By (2) we deduce $y>z$. Then $f(y,z)>f(z,y)$ by (1),but $f(y,z)=f(x,y)$ by hypothesis, so $f(x,y)>f(z,y)$. By (2) we deduce $x>z$. 
At this point we have $y>x>z$. Then $f(x,z)>f(z,x)$ by (1),but $f(z,x)=f(y,z)$ by hypothesis, so $f(x,z)>f(y,z)$ which contradicts (1).
This shows $x\geq y$. Similar arguments show that $y\geq z$ and $z\geq x$, and finally $x=y=z$.
