2017 AMC 10B Problem 3

P1-10~2 minPrint

Absolute Value & Inequalities Bound the quantity above/below or estimate to pin it down Test small/specific values or special cases to find or verify the answer

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?

Show solution without answering

Key insight

Add the two facts: y is more than -1 and z is more than 1, so y + z is more than 0. Every other choice can be negative.

Solution

"Necessarily positive" means positive for every allowed choice of $x$ , $y$ , $z$ . So a choice is right only if we can give a reason that always works, and a choice is wrong as soon as we find one example where it is not positive.

Choice (E) always works. We know $y > -1$ and $z > 1$ . Adding these gives $y + z > -1 + 1 = 0$ . So $y + z$ is always positive.
The others can fail. The only negative number is $y$ , so try to make $y$ win. One example for each is enough:

Choice	Numbers used	Value
(A) $y + x^2$	$x = 0.1$ , $y = -0.9$	$-0.9 + 0.01 = -0.89$
(B) $y + xz$	$x = 0.1$ , $y = -0.9$ , $z = 1.5$	$-0.9 + 0.15 = -0.75$
(C) $y + y^2$	$y = -0.5$	$-0.5 + 0.25 = -0.25$
(D) $y + 2y^2$	$y = -0.25$	$-0.25 + 0.125 = -0.125$

All four values are negative, so only (E) is always positive.

The answer is $\boxed{\textbf{(E)}\ y+z}$ .

Why this works

"Always true" needs a reason; "not always true" needs just one example. For a number between $-1$ and $0$ , its square is positive but smaller in size than the number itself (for example $(-0.5)^2 = 0.25$ ), so adding the square is not enough to make the total positive.

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

Choice (E): from $y > -1$ and $z > 1$ , adding gives $y + z > 0$ . So (E) is always positive.

To be sure the others fail, produce one counterexample for each. Take $y$ near $-1$ where it hurts most:

(A) $x = 0.1$ , $y = -0.9$ : $y + x^2 = -0.89 < 0$ .
(B) $x = 0.1$ , $z = 1.5$ , $y = -0.9$ : $y + xz = -0.75 < 0$ .
(C) $y = -0.5$ : $y + y^2 = -0.25 < 0$ .
(D) $y = -0.25$ : $y + 2y^2 = -0.25 + 0.125 < 0$ .

The answer is $\boxed{\textbf{(E)}\ y+z}$ .

Review in the AoPS books

Introduction to Algebra, Chapter 20: Special Functions (§20.2, §20.5)
Introduction to Algebra, Chapter 15: More Inequalities (§15.1, §15.3, §15.4)
Intermediate Algebra, Chapter 12: Inequalities (§12.3, §12.4)