A proof for $0 \le k \le 5$.
Assume that $0 \le k \le 5$.
We need to prove that, for all $a, b, c > 0$ with $abc = 1$,
$$\sum_{\mathrm{cyc}} \left(\sqrt{\frac{a^6 + k}{a(a^3+k)}} - a\right) \ge 0. \tag{1}$$
Using $\ln a + \ln b + \ln c = \ln(abc) = 0$, (1) is written as
$$\sum_{\mathrm{cyc}} \left(\sqrt{\frac{a^6 + k}{a(a^3+k)}} - a + \frac{3k}{2k+2}\ln a\right) \ge 0. \tag{2}$$
We can prove that, for all $x > 0$,
$$\sqrt{\frac{x^6 + k}{x(x^3+k)}} - x + \frac{3k}{2k+2}\ln x \ge 0. \tag{3}$$
(The proof of (3) is given at the end.)
Thus, (2) is true.
$\phantom{2}$
Remark. For $5 < k \le 12$, Perhaps it is difficult to tackle this inequality by the Olympiad style approaches. We may use the method of Lagrange Multipliers. The approach is similar to my answer here.
Let
$$f(x) := \sqrt{\frac{x^6 + k}{x(x^3+k)}} - x.$$
We need to prove that, for all $a, b, c > 0$ with $abc = 1$,
$$f(a) + f(b) + f(c) \ge 0.$$
Consider the minimum of $f(a) + f(b) + f(c)$ subject to $a, b, c > 0$ and $abc = 1$.
It is easy to prove that the minimum occurs (finite), and furthermore, the minimizer lies in $a, b, c \in [0, M]$ for some $M > 0$.
The method of Lagrange Multipliers yields the system of equations
\begin{align*}
f'(a) &= \lambda bc, \\
f'(b) &= \lambda ca, \\
f'(c) &= \lambda ab,\\
abc &= 1.\tag{1}
\end{align*}
Clearly, we have $af'(a) = bf'(b) = cf'(c) = \lambda$.
I think we can try to prove that if $a, b, c > 0$ and $\lambda$ satisfy the system of equations (1), then $a = b = c = 1$.
$\phantom{2}$
Proof of (3).
We have
$$\sqrt{\frac{x^6 + k}{x(x^3+k)}} - x
= \frac{\frac{x^6 + k}{x(x^3+k)} - x^2}{\sqrt{\frac{x^6 + k}{x(x^3+k)}} + x}
= \frac{k(1-x)(x^2 + x + 1)}{x(x^3 + k)\left(\sqrt{\frac{x^6 + k}{x(x^3+k)}} + x\right)}$$
$$\ge \frac{k(1-x)(x^2 + x + 1)}{x(x^3 + k)\left(\sqrt{\frac{1}{x}} + x\right)}$$
where we use $\frac{x^6 + k}{x^3+k} \le 1$ for all $0 < x \le 1$,
and $\frac{x^6 + k}{x^3+k} > 1$ for all $x > 1$.
It suffices to prove that
$$\frac{k(1-x)(x^2 + x + 1)}{x(x^3 + k)\left(\sqrt{\frac{1}{x}} + x\right)} + \frac{3k}{2k+2} \ln x \ge 0,$$
or
$$\frac{x^2 + x + 1}{x\left(\sqrt{\frac{1}{x}} + x\right)}\cdot (1-x)\frac{2k + 2}{x^3 + k} + 3 \ln x \ge 0,$$
or (using $(1-x)\frac{2k+2}{x^3 + k} \ge (1-x)\frac{2\cdot 5+2}{x^3 + 5}$ for all $x > 0$)
$$\frac{x^2 + x + 1}{x\left(\sqrt{\frac{1}{x}} + x\right)}\cdot (1-x)\frac{2\cdot 5 + 2}{x^3 + 5} + 3 \ln x \ge 0.$$
If $x \ge 1$, using $\ln u \ge \frac{2(u-1)}{u+1}$ for all $u\ge 1$ (easy), it suffices to prove that
$$\frac{x^2 + x + 1}{x\left(\sqrt{\frac{1}{x}} + x\right)}\cdot (1-x)\frac{2\cdot 5 + 2}{x^3 + 5} + 3 \cdot \frac{2(x-1)}{x+1} \ge 0, $$
which is true.
If $0 < x < 1$, using
$\ln x \ge \frac{1}{\sqrt{x}}\frac{(x-1)(17x^2 + 206x + 17)}{3(9x^2 + 62x + 9)}$ for all $0 < x < 1$ (taking derivative), it suffices to prove that
$$\frac{x^2 + x + 1}{x\left(\sqrt{\frac{1}{x}} + x\right)}\cdot (1-x)\frac{2\cdot 5 + 2}{x^3 + 5} + 3 \cdot \frac{1}{\sqrt{x}}\frac{(x-1)(17x^2 + 206x + 17)}{3(9x^2 + 62x + 9)} \ge 0, $$
which is true.
We are done.
