Is the Inverse of F(x) = x/(1+x+y+z) Possible to Find?

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trying to find the inverse of z=x/(1+x+y+w)
 
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What are your thoughts on the matter?
 
this is the function I'm actually dealing with:

F=<f_1, f_2, f_3> , where f_k(x_1,x_2,x_3) = x_k/(1+x_1+x_2+x_3) for k=1,2,3 and where x_1+x_2+x_3 is not -1 for all x_1,x_2,x_3

using the inverse function theorem, i found the jacobian to be (1+x_1+x_2+x_3)^(-4) which is not zero, and since all the 1st order partial derivatives exist and are continuous, this makes me think that F has an inverse, but actually finding out what it is is where I'm getting stuck
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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