Question

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?

$y+x^2$

$y+xz$

$y+y^2$

$y+2y^2$

$y+z$

The choice $y+z$ can be tested directly from the given bounds

Use one small counterexample to reject each other expression

Since $-1 < y$ and $1 < z,$ we can add the inequalities to see that $0 < y+z.$ This naturally proves choice E correct.

Furthermore, we can eliminate every other choice with the following values: $x=0.1,$ $y=-0.25,$ $z=1.25.$

Thus, the correct answer is E .