Real numbers $x$, $y$, and $z$ satisfy the inequalities
$0<x<1$, $-1<y<0$, and $1<z<2$.
Which of the following numbers is necessarily positive?
$\textbf{(A)}\ y+x^2\qquad\textbf{(B)}\ y+xz\qquad\textbf{(C)}\ y+y^2\qquad\textbf{(D)}\ y+2y^2\qquad\textbf{(E)}\ y+z$
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.