2. Given the three sides of $\triangle A B C$ are $a, b, c$. The following three conclusions:
(1) A triangle with side lengths $\sqrt{a}, \sqrt{b}, \sqrt{c}$ must exist;
(2) A triangle with side lengths $a^{2}, b^{2}, c^{2}$ must exist;
(3) A triangle with side lengths $|a-b|+1, |b-c|+1, |c-a|+1$ must exist.
Among them, the number of correct conclusions is ( ).
(A) 0
(B) 1
(C) 2
(D) 3
2. C.

Let $a \geqslant b \geqslant c$. Then $b+c>a$.
(1) Since $b+c>a$, we have
$$
b+c+2 \sqrt{b c}>a \Rightarrow \sqrt{b}+\sqrt{c}>\sqrt{a} \text {. }
$$

Therefore, a triangle with side lengths $\sqrt{a}, \sqrt{b}, \sqrt{c}$ must exist.
(2) A triangle can be formed with side lengths $a=2, b=3, c=4$, but a triangle with side lengths $a^{2}=4, b^{2}=9, c^{2}=16$ does not exist.
(3) Since $a \geqslant b \geqslant c$, we have
$$
\begin{array}{l}
|a-b|+1=a-b+1, \\
|b-c|+1=b-c+1, \\
|c-a|+1=a-c+1 .
\end{array}
$$

Thus, among the three sides, $|c-a|+1$ is not less than the other two sides.
$$
\begin{array}{l}
\text { and }(|a-b|+1)+(|b-c|+1) \\
\geqslant|a-c|+1+1 \\
>|c-a|+1,
\end{array}
$$

Therefore, a triangle with side lengths $|a-b|+1, |b-c|+1, |c-a|+1$ must exist.