Classical Inequalities addicts can do this. Part 5

$\begin{cases} \begin{aligned} a^2 + b^2 + c^2 &= 2016 \\ a + 2b + 3c &= k \end{aligned} \end{cases}$

$\begin{cases} \begin{aligned} a^2 + c^2 &= 2016 - b^2 \\ a + 3c &= k - 2b \end{aligned} \end{cases}$

By Cauchy,
$\begin{aligned} (a+3c)^2 \leq 10(a^2 + c^2) \implies & (k-2b)^2 \leq 10(2016-b^2) \\ &k^2 - 4kb + 4b^2 \leq 20160-10b^2 \\ &14b^2 - 4kb + (k^2 - 20160) \leq 0\end{aligned}$

By quadratic formula, we have: $\dfrac{4k - \sqrt{1128960 - 40k^2}}{28} \leq b \leq \dfrac{4k + \sqrt{1128960 - 40k^2}}{28} \cdot$

Note that: $1128960 - 40k^2 \geq 0 \implies -168 \leq k \leq 168 \cdot$

Trying $k = 168$ yields $14(b - 24)^2 \leq 0 \implies b \leq 24 \implies \max(b) = 24$ which is an integer, which means, $\max(k) = 168$ will suffice the condition above.