Show that two random variables are independent I'm learning probability and need help with the following problem : 

Let $X_1, X_2$ be independent and identically distributed random variables with probability density function 
$$f(x_i) = \begin{cases} \lambda_i e^{\lambda_i x_i}, & x_i > 0 \;\; (i = 1, 2) \\ 0 & \text{elsewhere}.\end{cases}$$
$(1)$ Find the distribution of the random variable $V = \min(X_1, X_2)$.
$(2)$ Show that the random variables $Z = X_1/(X_1+X_2)$ and $X_1+X_2$ are independent. 

Since I'm having difficulties for $(2)$, I'm going to share my work for $(1)$.
$(1)$ If two variables are iid, then they must have the same distribution. Which means they have the same parameters. So I assumed $\lambda_1 = \lambda_2$, i.e. 
$$X_1, X_2 \overset{idd}\sim exp(rate = \lambda).$$ 
To find the distribution of the minimum, we look at cdfs. The cdf for the individual $X_i$’s is
$$F(x)  = \int_{-\infty}^{x} f(u) \, du = \int_{0}^{x} f(u) \, du =  \int_{0}^{x} \lambda e^{-\lambda u} \, du = 1 - e^{-\lambda x}.$$
The cdf for $V$, the minimum, is :
\begin{align*} F_V(v) = P(V \leq v) 
&= P(\min(X_1, X_2) \leq v) \\ \\
&= 1 - P(\min(X_1, X_2) > v) \\ \\
&\overset{indep}= 1 - P(X_1 > v) \cdot P(X_2 > v) \\ \\
&\overset{ident}= 1 - [P(X_1 > v)]^n. \end{align*}
From are computed cdf, we have that 
$$P(X_1 > v) = 1 - P(X_1 \leq v) = 1 - (1 - e^{-\lambda v}) = e^{-\lambda v}.$$
So, the cdf for the minimum $V$ is 
$$F_V(v) = 1 - [e^{-\lambda v}]^2 = 1 - e^{-\lambda 2 v}.$$
The pdf for the minimum is 
$$f_V(v) = F'_V(v) = 2\lambda e^{-2\lambda v}.$$
So we proved that $$V \sim exp(rate  = 2\lambda).$$

Is my work correct for $(1)$? Unfortunately I have no idea how to solve $(2)$. Any help would be appreciated. 
 A: $(1)$ The variables $X_1$, $X_2\sim \mathcal{E}(\lambda)$ are i.i.d. exponentially distributed. As you showed via the distribution function method, the minimum is exponentially distributed too, and the parameter for the minimum is $2\lambda$, i.e. the sum of the parameters of $X_1$ and $X_2$ (see e.g. [1]).
$(2)$ Since $X_1$, $X_2$ are independent, the joint distribution probability distribution function (PDF) of the vector $X = (X_1, X_2)$ writes:
$$
f_X(x) = \lambda^2 \exp\left(-\lambda (x_1 + x_2)\right)\,\boldsymbol{1}_{x_1\geq 0,x_2\geq 0} \, .
$$
To show that the components $Y_1 = X_1/(X_1 + X_2)$ and $Y_2 = X_1 + X_2$ of the vector $Y = g(X)$ are independent, one can show that the joint PDF $f_Y$ of $Y$ is separable. To do so, let us compute $f_Y$ by the change of variable method. We consider a bounded measurable function $h$, and compute $\mathbb{E}h(Y)$:
\begin{aligned}
\mathbb{E}(h\circ g)(X) &= \int_{\mathbb{R}^4} h\circ g(x)\, f_{X}(x) \,\mathrm{d}x \, ,\\
&= \int_{\mathbb{R}^4} h(y)\, \underbrace{f_{X}\big({g}^{-1}(y)\big) \big|J_{{g}^{-1}}(y)\big|}_{f_Y(y)} \,\mathrm{d}y \, ,
\end{aligned}
where $J_{g^{-1}}$ is the Jacobian of the reciprocal function of $g$, defined by $$ g^{-1}(y) = \big(y_1 y_2 , y_2 (1-y_1)\big) . $$
We get
$$
f_Y(y) = \lambda^2\, {y_2} \exp\left(-\lambda y_2 \right)\, \boldsymbol{1}_{0\leq y_2}\, \boldsymbol{1}_{0\leq y_1\leq 1}  \, ,
$$
which is in separated form. Note that this question is similar to [2, 3, 4].
