What we are given is

\frac{1}{X}>Y+Z

Now, for (A), if we have X=-1, then

\frac{1}{X}>Y+Z

becomes

\frac{1}{-1}>Y+Z

or equivalently,

-1>Y+Z

So, since we know that this inequality is true for sure, we know that and number greater than or equal to -1 will be greater that Y+Z.

And for X>Y+Z, we are comparing X=-1 to Y+Z, which we already know that Y+Z is always less than -1, and hence X=-1 is greater than Y+Z.

If this confused you, just think of what you are given, that \frac{1}{X}=\frac{1}{-1}=-1>Y+Z, and what can you conclude from that.

How to find that the case of X=-1 satisfies the new inequality? Well, you can think that X and \frac{1}{X} are equal for X equal 1 or -1, so if \frac{1}{X} is greater than Y+Z, then so is the value X since \frac{1}{X}=X. And since we know that X<0, we exclude the case of X=1.

\textbf{EXTRA:}

Indeed, we can find more numbers than -1 that satisfy both inequalities.

To have X>Y+Z given that \frac{1}{X}>Y+Z, we should require that

X\ge\frac{1}{X}>Y+Z

(If this does not make sense to you, think if we have 5>a+b then to find another number that is greater than a+b, we should require it to be greater than or equal 5).

So, we need

X\ge\frac{1}{X}

Multiplying both sides by X while keeping in mind that X<0, we get

X^2<1\qquad(*)

Thus, for any negative X that satisfies (*), we get that both X and \frac{1}{X} are greater than Y+Z. You can try taking X=-\frac{1}{2} for example!