I understand that i.i.d. r.v.s are independent and have the same probability distribution, but I don't understand why that implies $$E\left(\frac X{X+Y+Z}\right)=E\left(\frac Y{X+Y+Z}\right)=E\left(\frac Z{X+Y+Z}\right)$$
The original problem was:
Let $X, Y, Z$ are positive i.i.d. r.v.s, find $E\left(\dfrac X{X+Y+Z}\right)$
The solution contained the aforementioned result, and at the end concluded that $$E\left(\frac X{X+Y+Z}\right)=\frac13$$
A simple explanation, please! :)
By symmetry, the random variables $U = \frac{X}{X+Y+Z}$, $V = \frac{Y}{X+Y+Z}$ and $W = \frac{Z}{X+Y+Z}$ have the same probability distribution since $(X,Y,Z)$, $(Y,Z,X)$ and $(Z,X,Y)$ are identically distributed. Hence $U,V,W$ have same expectation.
However some hypotheses are missing here, for instance $X,Y,Z$ with nonzero positive values. Otherwise it might happen that $U,V,W$ are not well-defined (if a division by zero occurs) or don't have a finite expectation.