$X$ is an exponential random variable of parameter λ when its probability distribution function is
$$f(x) = \begin{cases}\lambda e^{-\lambda x} & x \geq 0 \\ 0 & x < 0 \end{cases}$$
For a > 0, we have the cumulative distribution function
$$F_x(a) = \int_0^a f(x) dx = \int_0^a \lambda e^{-\lambda x} dx = -e^{-\lambda x} \mid_0^a = 1 - e^ {-\lambda a}$$
So,
$$P\left\{X < a \right \} = 1 - e^ {-\lambda a} $$ and
$$P\left\{X > a \right \} = e^ {-\lambda a} $$
Now, we use $P \left \{X > a \right \}$ for our problem because our concerned variable $Z$ is min of $X$ and $Y$.
For exponential distribution with parameter $\lambda$, mean is given by $\frac{1}{\lambda}$.
We have,
$P \left \{X > a \right \} = e^ {-\frac{1}{\alpha} a} $
$P \left \{Y > a \right \} = e^ {-\frac{1}{\beta} a} $
So, $\begin{align*}P\left \{Z > a \right \} &= P \left \{X > a \right \} P \left \{Y > a \right \} \left(\because \text{X and Y are independent events and } \\Z > \min \left(X, Y \right) \right)\\&=e^ {-\frac{1}{\alpha} a} e^ {-\frac{1}{\beta} a} \\&=e^{-\left(\frac{1}{\alpha} + \frac{1}{\beta} \right)a} \\&=e^{-\left(\frac{\alpha + \beta} {\alpha \beta} \right)a}\end{align*}$
This shows that $Z$ is also exponentially distributed with parameter $\frac{\alpha + \beta} {\alpha \beta}$ and mean $\frac{\alpha \beta} {\alpha + \beta}$.
Ref: http://ocw.mit.edu/courses/mathematics/18-440-probability-and-random-variables-spring-2011/lecture-notes/MIT18_440S11_Lecture20.pdf