Statement 1: u+v > 0If u(u+v) ≠0 and u > 0, is 1/(u+v) < 1/u + v?
1) u+v >0
2) v>0
Plugging u=1 and v=1 into 1/(u+v) < 1/u + v, we get:
1/(1+1) < 1/1 + 1
1/2 < 2.
YES.
Plugging u=2 and v=-1 into 1/(u+v) < 1/u + v, we get:
1/(2-1) < 1/2 - 1
1 < -1/2.
NO.
Since in the first case the answer is YES, but in the second case the answer is NO, INSUFFICIENT.
Statement 2: v>0
Since all of the values are positive, we can rephrase the question stem.
Is 1/(u+v) < 1/u + v?
Put the right side over a common denominator:
1/(u+v) < (1+uv)/u
Cross-multiply:
u < (u+v)(1+uv)
Distribute on the right-hand side:
u < u + u²v + v + uv².
Subtract u from each side:
0 < u²v + v + uv².
Question stem, rephrased:
Is u²v + v + uv² > 0?
Since all of the values on the left-hand side are positive, the answer is YES.
SUFFICIENT.
The correct answer is B.
