Let's start with the stem:

√(xy) > (xy)²

√(xy) - (xy)² > 0

√(xy) * (1 - (xy)¹ ̇�) > 0

Since we know xy > 0, we must have √(xy) > 0. The question then becomes

Is (1 - (xy)¹ ̇�) > 0 ?

or

Is 1 > (xy)¹ ̇� ?

or

Is 1 > xy * (√xy) ?

or

Is 1 > xy > 0 ?

S1:

If x > 0, then we have 4 > 7xy, or (4/7) > xy. In that case, we're set: 1 > 4/7 > xy > 0.

If 0 > x, then we have 4 < 7xy, or (4/7) < xy. In this case, we're NOT set: we could have x = -3 and y = -10, or whatever.

S2:

x > 0

By itself this isn't helpful.

S1 + S2

x > 0, so only the first case mentioned in S1 applies.

Mo2men wrote:If xy > 0, is (xy)^2 < √(xy)

(1) 4/x > 7y
(2) x − 16 > −16

Note that the inequality in the question prompt will be true for values of xy, such that 0 < xy < 1. So the question can be rephrased as "Is 0<xy<1?"

(1) NOT SUFFICIENT. If x > 0, the inequality becomes 4/7 > xy. Which combined with the restriction that xy>0 would be an affirmative answer to the prompt. However, if x<0, then the inequality becomes 4/7 < xy. Which is insufficient.

Here are two examples: 4/-2 > 7(-3). xy = (-2)(-3) = 6. 36 > √(6).

4/(1/2) > 7(1). xy = (1/2)(1) = 1/2. 1/4 < √(1/2).

(2) NOT SUFFICIENT. x>0. Tells us nothing about y.

(1) & (2) Combined. SUFFICIENT. (2) eliminates the case in (1) where x<0. Hence, 0<xy<4/7 & the prompt question must be true.

Answer choice: C

Another way to think about this:

Let's say that xy = z, so we're dealing with a single variable.

Now our question is "Is √z > z²", which makes us think of values of z that might satisfy this inequality.

If z > 1, then z² > √z, since z² makes z bigger and √z makes z smaller. But if 1 > z > 0, we're in this twilight zone of positive numbers where √z makes z BIGGER and z² makes z SMALLER. This is our range, so our question becomes

"Is 1 > z > 0" ?

or, since z was our dummy variable that stood for xy

"Is 1 > xy > 0" ?

From there, we can use the approach detailed in my first solution (but without a lot of the technical algebra in the setup!)