What would be the ${Corr}(X,Y+Z)$ equal to, given Y,Z are independent random variables I know that ${corr}(X,Y) = \frac{cov(X,Y)}{\sigma_x\sigma_y}$
a
for ${corr}(X,Y+Z) = \frac{cov(X,Y+Z)}{\sigma_x\sigma_{y+z}}$
$$cov(X,Y+Z) = E[X(Y+Z)] - E[X]E[Y+Z] \\
 cov(X,Y+Z)=E[XY] + E[XZ] - E[X]E[Y] - E[X]E[Z] \\
cov(X,Y+Z) = cov(X,Y) + cov(X,Z)
$$
$\sigma_{y+z} = \sqrt{\sigma_y + \sigma_z} $
finally
$corr(X,Y+Z) = \frac{cov(X,Y) + cov(X,Z)}{ \sigma_x \sqrt{\sigma_y + \sigma_z}}$ now how do i simplify this in terms of corr(X,Y) and corr(X,Z) ? 
 A: Picking up from where you left off (modulo J.G.'s correct observation in the comments):
\begin{align*}
\operatorname{corr}(X, Y + Z) &= \frac{\operatorname{cov}(X, Y)}{\sigma_x} \cdot \frac{1}{\sqrt{\sigma_y^2 + \sigma_z^2}} + \frac{\operatorname{cov}(X, Z)}{\sigma_x} \cdot \frac{1}{\sqrt{\sigma_y^2 + \sigma_z^2}} \\
  &= \frac{\operatorname{cov}(X, Y)}{\sigma_x\sigma_y} \cdot \frac{\sigma_y}{\sqrt{\sigma_y^2 + \sigma_z^2}} + \frac{\operatorname{cov}(X, Z)}{\sigma_x \sigma_z} \cdot \frac{\sigma_z}{\sqrt{\sigma_y^2 + \sigma_z^2}} \\
  &= \operatorname{corr}(X, Y) \cdot \frac{1}{\sqrt{1 + (\frac{\sigma_z}{\sigma_y})^2}} + \operatorname{corr}(X, Z) \cdot \frac{1}{\sqrt{1 + (\frac{\sigma_y}{\sigma_z})^2}}
\end{align*}
Note the indices in the squared term on the last line, since they're small; the term in parentheses under the radical in the left denominator is $\dfrac{\sigma_z}{\sigma_y}$, and the corresponding item on the right is its reciprocal.
It might be possible to simplify this in some way, but I don't think it will be possible to avoid using $\sigma_y$ and $\sigma_z$ in any equivalent expression. Of course, if you knew something like $\sigma_z = \sigma_y$ (for example), you could clean this up quite nicely.
