(A) for me too

(x^-1 + y^-1)^-1 > [(x^-1)(y^-1)]^-1 ?

<=> 1/[1/x + 1/y] > 1/(1/x * 1/y) ?

=> x*y/(x+y) > x*y ? (as we know that x != 0 and y != 0)

<=> x*y/(x+y) - x*y > 0 ?

<=> x*y * [1/(x+y) - 1] > 0 ?

From 1
x = 2y,

We have so,

x*y * [1/(x+y) - 1]

= 2*y * y *[1/(2*y + y) -1]

= 2*y *[ 1/3 - y]

We have 2 cases:

o If y < 0, then 2*y < 0 and 1/3 - y > 0 as y =< -1 (integer)... thus 2*y *[ 1/3 - y] < 0

o If y > 0, then 2*y > 0 and 1/3 - y < 0 as y >= 1 (integer)... thus 2*y *[ 1/3 - y] < 0

SUFF.

From 2
x + y > 0

Implies that:

1/(x+y) - 1 =< 0 because x and y are integers so x+y >= 1

Thus,

o If x+y=1 (x=3 and y=-2), then x*y * [1/(x+y) - 1] = 0

o If x+y > 1 and x > 0 & y > 0 (x=3 and y=3), then x*y * [1/(x+y) - 1] < 0.

INSUFF.