3. Given that $a$, $b$, and $c$ are positive numbers, and
$$
\frac{b^{2}+c^{2}-a^{2}}{2 b c}+\frac{a^{2}+c^{2}-b^{2}}{2 a c}+\frac{a^{2}+b^{2}-c^{2}}{2 a b}>1 \text{. }
$$

Then, with $a$, $b$, and $c$ as sides, a triangle can be formed (fill in "can" or "cannot").
3. Can.

Given that $a$, $b$, and $c$ are positive numbers, removing the denominators from the given equation yields
$$
\begin{array}{l}
a\left(b^{2}+c^{2}-a^{2}\right)+b\left(a^{2}+c^{2}-b^{2}\right) \\
>2 a b c-c\left(a^{2}+b^{2}-c^{2}\right) .
\end{array}
$$

Factoring both sides of the above equation, we get
$$
\begin{array}{l}
(a+b)(a+c-b)(b+c-a) \\
>c(a+c-b)(b+c-a),
\end{array}
$$

which simplifies to $(a+b-c)(a+c-b)(b+c-a)>0$.
By symmetry, without loss of generality, assume $a \geqslant b \geqslant c$. Then
$$
a+b-c>0, a+c-b>0 \text {. }
$$

Therefore, $b+c-a>0$, which means $b+c>a$.
Hence, $a$, $b$, and $c$ can form a triangle.