3. Let $a, b, c$ be positive numbers, and
$$
a^{2}+b^{2}+c^{2}-2 a b-2 a c-2 b c<0 \text {. }
$$

Then with $\sqrt{a}, \sqrt{b}, \sqrt{c}$ as sides \qquad a triangle can be formed (fill in “can” or “cannot”).

3. Can.

Assume $a \geqslant b \geqslant c$. Because
$$
\begin{array}{l} 
a^{2}+b^{2}+c^{2}-2 a b-2 a c-2 b c \\
= {\left[\left(a^{2}+2 a b+b^{2}\right)-2(a c+b c)+c^{2}\right]-4 a b } \\
=(a+b-c)^{2}-(2 \sqrt{a b})^{2} \\
=(a+b-c+2 \sqrt{a b})(a+b-c-2 \sqrt{a b}) \\
= {\left[(\sqrt{a}+\sqrt{b})^{2}-(\sqrt{c})^{2}\right]\left[(\sqrt{a}-\sqrt{b})^{2}-(\sqrt{c})^{2}\right] } \\
=(\sqrt{a}+\sqrt{b}+\sqrt{c})(\sqrt{a}+\sqrt{b}-\sqrt{c}) . \\
(\sqrt{a}-\sqrt{b}+\sqrt{c})(\sqrt{a}-\sqrt{b}-\sqrt{c}) \\
0, \sqrt{a}+\sqrt{b}-\sqrt{c}>0$, $\sqrt{a}-\sqrt{b}+\sqrt{c}>0$,
Therefore, $\sqrt{a}-\sqrt{b}-\sqrt{c}\sqrt{a}$.
Hence, a triangle can be formed with sides $\sqrt{a} 、 \sqrt{b} 、 \sqrt{c}$.