# 2017 AMC 10B Problems/Problem 3

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## Problem

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

## Solution

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, $|y|>|y^2|$, and thus it is always negative. $\textbf{(D)}$ The same logic as above, but when $-\frac{1}{2} this time.

~savannahsolver

## Video Solution by TheBeautyofMath

~IceMatrix

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 