smc

Notice that $y+z$ must be positive because $|z|>|y|$. Therefore the answer is $\boxed{y+z}$. The other choices: $\text{(A)}$ As $x$ grows closer to $0$, $x^2$ decreases and thus becomes less than $y$. $\text{(B)}$ $x$ can be as small as possible ($x>0$), so $xz$ grows close to $0$ as $x$ approaches $0$. $\text{(C)}$ For all $-1<y<0$, $|y|>|y^2|$, and thus it is always negative. $\text{(D)}$ The same logic as above, but when $-\frac{1}{2}<y<0$ this time.