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By Cauchy-Schwarz, $\begin{aligned} 4(b^2+c^2+d^2+e^2)&\geq (b+c+d+e)^2\\ 4(16-a^2)&\geq (8-a)^2\\ a(5a-16)&\leq 0, \end{aligned}$ and thus $0\leq a\leq \dfrac{16}{5}.$ Such a maximum can be achieved when $b=c=d=e=\tfrac 65$ .

Hence, the largest possible value of $a$ is $\tfrac{16}{5}=\boxed{3.2}$ .

Given that $a^2+b^2+c^2+d^2+e^2 = 16$ , $a$ is maximum, when $a^2$ is also maximum or $b^2+c^2+d^2+e^2$ is minimum. By Titu's lemma , we have $\dfrac {b^2}1+\dfrac {c^2}1+\dfrac {d^2}1+\dfrac {e^2}1 \le \dfrac {(b+c+d+e)^2}4$ . Since equality occurs when $b=c=d=e$ , the system of equations becomes.

$\begin{cases} a+b+c+d+e = a+4b = 8 & ...(1) \\ a^2+b^2+c^2+d^2+e^2 = a^2 + 4b^2 = 16 & ...(2) \end{cases}$

$\begin{aligned} (2): \qquad \qquad {\color{#3D99F6} a^2} + 4b^2 & = 16 & \small \color{#3D99F6} \text{From }(1): \ a = 8 - 4b \\ {\color{#3D99F6} 64-64b + 16b^2} + 4b^2 & = 16 \\ 5b^2 - 16b + 12 & = 0 \\ (5b-6)(b-2) & = 0 \\ \implies b & = \frac 65 & \small \color{#3D99F6} \text{for minimum }b^2+c^2+d^2+e^2=4b^2 \end{aligned}$

Therefore, $\max(a) = 8 - 4\min(b) = 8 - 4 \times \dfrac 65 = \dfrac {16}5 = \boxed{3.2}$ .

Note that there is no solution if any, some or all of $b$ , $c$ , $d$ , and $e$ are $\le 0$ .

[Not complete solution]

$QM-AM$ $\sqrt{\frac{b^{2}+c^{2}+d^{2}+e^{2}}{4}} \geq \frac{b+c+d+e}{4}$ then substitute $a$ and we done.