Answer
Sometimes
Work Step by Step
Let $x=1, y=1$.
$x+y > 0$
$1+1 > 0$
$2 > 0$ (true)
$x*y> 0 $
$1*1 > 0$
$1 > 0$ (true)
Let $x=-1, y=2$
$x+y > 0$
$-1+2 > 0$
$ 1 > 0$ (true)
$x*y > 0$
$-1*2 > 0$
$-2 > 0$ (false)
Let $x=-2, y=1$
$x+y > 0$
$-2 + 1 > 0$
$-1 > 0$ (false)
$x*y > 0$
$-2*1 > 0$
$-2 > 0$ (false)
Let $x=-2, y=-3$
$x+y > 0$
$-2 + (-3) > 0$
$-5 > 0$ (false)
$x*y > 0$
$(-2)*(-3) > 0$
$6 > 0$ (true)
The statement is sometimes true. If we let $x>0$ and $y>0$, then the statement is always true. If we let $abs(x) > abs(y) > 0$ (with either variable negative), then only part of the statement is true. If both variables are negative, then only part of the statement is true.
