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