If x and y are nonzero integers, is (x^(-1) + y^(-1))^(-1) >

First of all the question should be:

If x and y are nonzero integers, is (x^(-1) + y^(-1))^(-1) > (x^(-1)*y^(-1))^(-1) ?

Is \((x^{-1}+y^{-1})^{-1}> (x^{-1}*y^{-1})^{-1}\)?

Is \((\frac{1}{x}+\frac{1}{y})^{-1}>(\frac{1}{xy})^{-1}\)?

Is \((\frac{x+y}{xy})^{-1}>xy\)?

Is \(\frac{xy}{x+y}>xy\)?

Now, from this point you cannot divide both parts of the inequality by \(xy\) and write \(\frac{1}{x+y}>1\) (as you did), because you don't know whether \(xy\) is positive or negative: if \(xy>0\) then you should write \(\frac{1}{x+y}>1\) BUT if \(xy<0\) then you should flip the sign and write \(\frac{1}{x+y}<1\). But even if you knew that \(xy>0\) then the next step of writing \(x+y<1\) from \(\frac{1}{x+y}>1\) would still be incorrect for the same exact reason: you don't k now whether \(x+y\) is positive or negative, hence you cannot multiply both sides of the inequality by \(x+y\).

Never multiply or divide inequality by a variable (or by an expression with variable) unless you are sure of its sign since you do not know whether you must flip the sign of the inequality.

Thus the question is boiled down to: is \(\frac{xy}{x+y}>xy\)? Actually we can manipulate further but there is no need.

(1) x = 2y --> question becomes: is \(\frac{2y^2}{3y}>2y^2\)? Now, as we know that \(y\) is nonzero then \(2y^2>0\) and we can divide both parts by it --> is \(\frac{1}{3y}>1\)? As \(y\) is an integer (no matter positive or negative) then the answer to this question is always NO (if it's a positive integer then \(\frac{1}{3y}<1\) and if it's a negative integer then again: \(\frac{1}{3y}<0<1\)). Sufficient.

(2) x + y > 0 --> if \(x=y=1\) then the answer will be NO but if \(x=3\) and \(y=-1\) then the answer will be YES. Not sufficient.

Answer: A.