Five. (20 points) Given
$$
f(x)=\frac{x^{4}+k x^{2}+1}{x^{4}+x^{2}+1}, k, x \in \mathbf{R} \text {. }
$$
(1) Find the maximum and minimum values of $f(x)$;
(2) Find all real numbers $k$ such that for any three real numbers $a, b, c$, there exists a triangle with side lengths $f(a)$, $f(b)$, and $f(c)$.
Five, $f(x)=1+\frac{(k-1) x^{2}}{x^{4}+x^{2}+1}$.
(1) Since $x^{4}+1 \geqslant 2 x^{2}$. Then $x^{4}+x^{2}+1 \geqslant 3 x^{2}$. Therefore,
$0 \leqslant \frac{x^{2}}{x^{4}+x^{2}+1} \leqslant \frac{1}{3}$.
When $k \geqslant 1$, $f(x)_{\text {max }}=\frac{k+2}{3}, f(x)_{\text {min }}=1$;
When $k<1$, $f(x)_{\text {max }}=1, f(x)_{\text {min }}=\frac{k+2}{3}$.
When $k \geqslant 1$, as long as $2 \times 1>\frac{k+2}{3}$, i.e., $1 \leqslant k<4$;
When $k<1$, as long as $2 \times \frac{k+2}{3}>1$, i.e., $-\frac{1}{2}<k<1$.
Therefore, $-\frac{1}{2}<k<4$ is the solution.