You will be amazed with this

Without loss of generality we will isolate $a$ and determine its maximum and minimum value using Cauchy's Inequality . With the $b_{i}$ 's all equalling $1$ we have that

$(b^{2} + c^{2} + d^{2} + e^{2})\left(\displaystyle\sum_{k=2}^{5} 1\right) \ge (b + c + d + e)^{2}$

$\Longrightarrow (16 - a^{2})*4 \ge (8 - a)^{2}$

$\Longrightarrow 64 - 4a^{2} \ge 64 - 16a + a^{2}$

$\Longrightarrow a(16 - 5a) \ge 0 \Longrightarrow 0 \le a \le \dfrac{16}{5}.$

The minimum value of $a = 0$ can be accommodated by setting $b = c = d = e = 2$ , and the maximum value of $a = \dfrac{16}{5}$ can be accommodated by setting $b = c = d = e = \dfrac{6}{5}.$

The sum of the minimum and maximum values of $a$ is then $0 + \dfrac{16}{5} = \dfrac{16}{5} = \boxed{3.2}.$

The first equation describes a (hyper)plane in 5-space perpendicular to the direction (1, 1, 1, 1, 1).

The second equation describes a sphere in 5-space.

The solutions of the system lie on the intersection of the 5-plane and the 5-sphere, which is a 4-sphere.

The points of minimum and maximum value of $a$ lie opposite each other on the 4-sphere. Their average position is the center of the 4-sphere, which lies in direction (1, 1, 1, 1, 1) from the origin. It is easy to see that its coordinates must be $(\tfrac85, \dots$ ).

Finally, the sum of minimum and maximum $a$ is twice the average, i.e. $\tfrac{16}5 = \boxed{3.2}$ .

The equation ${ a }^{ 2 } + { b }^{ 2 } + { c }^{ 2 } + { d }^{ 2 } + { e }^{ 2 } = 16$ can be written as

$\large\ { b }^{ 2 } + { c }^{ 2 } + { d }^{ 2 } + { e }^{ 2 } = 16 - { a}^{ 2 }$ and by R.M.S-A.M the L.H.S is greater than or equal to

$\large\ \frac { { (b + c + d + e) }^{ 2 } }{ 4 }$ and this is equal to $\frac { { (8 - { a }) }^{ 2 } }{ 4 }$ .

Hence, $\large\ 16 - { a }^{ 2 }= \frac { { (8 - { a }) }^{ 2 } }{ 4 }$

or $\large\ 16={ a }^{ 2 } + \frac { { (8 - { a }) }^{ 2 } }{ 4 }$

or $0 \ge 5{ a }^{ 2 } - 16a = a(5a - 16)$

Therefore, $\large\ 0 \le a \le \frac { 16 }{ 5 }$ , where $a = 0$ iff $b = c = d = e = 2$ and $\large\ a = \frac { 16 }{ 5 }$ iff $b = c = d = e = \frac { 6 }{ 5 }$ .

Thus, minimum + maximum = $\large\ \boxed{\frac { 16 }{ 5 }}$