Correlated stock prices and geometric Brownian motion

I have two uncorrelated stocks which follow geometric Brownian motion, as follows

\begin{aligned} dS_a &= \mu_aS_adt + \sigma_aS_adW\\ dS_b &= \mu_bS_bdt + \sigma_bS_b dW \end{aligned}

Does a portfolio of these stocks also follow geometric Brownian motion?

I have determined that

$$dS_a + dS_b = (\mu_aS_a + \mu_bS_b)dt + (\sigma_aS_a + \sigma_bS_b)dW$$

which does not follow geometric Brownian motion. I'm stuck now on what happens if the $$S_a$$ and $$S_b$$ are correlated? How does this change?

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You posed a quiet vague question but I will try to reply to it. Let $$(\Omega, \mathcal{F}, \mathbb{P})$$ be a probability space. We denote $$X$$ the portfolio of two stocks which follow geometric brownian motions, i.e. for all $$t \in \mathbb{R}^+$$, \begin{align*} X_t = S^a_t + S^b_t \end{align*} where $$S^a_t$$ and $$S^b_t$$ have the following $$SDE$$: \begin{align*} dS^{x}_t = \mu^x_tdt + \sigma^x_tdW^x_t \end{align*} where $$x = \lbrace{a,b\rbrace}$$, $$\mu^a \neq \mu^b$$ and $$\sigma^a \neq \sigma^b$$. Suppose that $$W_t^a$$ and $$W_t^b$$ are two correlated brownian motions, i.e. $$d_t = \rho dt$$. By applying the Ito formula to the function $$\phi(x,y) = x+y$$, we have : \begin{align*} dX_t &= dS_t^b + dS_t^a \\ &= (S_t^a\mu^a_t + S_t^b\mu^b_t)dt + S_t^a\sigma^a_tdW^a_t + S_t^b\sigma^b_tdW^b_t \end{align*} As you can see, we supposed that the portfolio is a linear function to the stocks (i.e. sum of the stocks). Thus, the correlation will not change anything as the second derivative of the function $$\phi$$ is zero. So the portfolio is not a geometric brownian motion.
Now, if we want to be more general we can suppose that the portfolio is a function of the stocks and smooth enough to apply the Ito formula (we can first suppose $$\mathcal{C}^2(\mathbb{R}_+^2,\mathbb{R}_+)$$. We have then: \begin{align*} dX_t = d\phi(S_t^a, S_t^b) = \partial_x\phi dS_t^a + \partial_y\phi dS_t^b + \frac12\left[\partial_{xx}\phi _t + \partial_{yy}\phi _t + 2\partial_{xy}\phi _t\right] \end{align*} Then we can try to find $$\phi$$ such that $$X$$ is a geometric BM.