3. Let $a>b>c, a+b+c=1$, and $a^{2}+b^{2}+c^{2}=1$. Then ( )
(A) $a+b>1$
(B) $a+b=1$
(C) $a+b<1$
(D) Cannot be determined, depends on the specific values of $a, b$
3. (A).

Take $a=\frac{1+\sqrt{5}}{4}, b=\frac{1}{2}, c=\frac{1-\sqrt{5}}{4}$, which can negate (B) and (C), but cannot negate (D). To determine (A), we can use proof by contradiction. Assume there exist $a_{0} 、 b_{0}$ such that $a_{0}+b_{0} \leqslant 1$, then there exists $c_{0} \geqslant 0$, and further $a_{0}>b_{v}>0$, we get
$$
\begin{aligned}
0 & 1$.
$$