3 Given that $a$ and $b$ are real numbers. Then $x+y>a+b$ and $x y>a b$ implies $x>a$ and $y>b$ ( ).
(A) Sufficient but not necessary condition
(B) Necessary but not sufficient condition
(C) Sufficient and necessary condition
(D) Neither sufficient nor necessary condition
3. (B).

When $a>0, b>0$, from $x>a, y>b$ we can deduce that $x+y > a+b$, and $xy > ab$.

However, the converse is not necessarily true. From $a>0, b>0, x+y>a+b, xy>ab$, it does not necessarily follow that $x>a, y>b$. For example, $x=1, y=\frac{1}{3}, a=\frac{1}{3}, b=\frac{3}{5}$ satisfies $1+\frac{1}{3}>\frac{1}{3}+\frac{3}{5}$, and $1 \cdot \frac{1}{3}>\frac{1}{3} \cdot \frac{3}{5}$. But $x=1>\frac{1}{3}=a, y=\frac{1}{3}<b$. Therefore, $x+y>a+b$ and $xy>ab$ is a necessary but not sufficient condition for $x>a$ and $y>b$.