Inequalities, absolute value, and sign charts
Tonight's idea
An inequality has a whole region of solutions. Three tools find it without missing a piece.
1. Absolute value is distance. \(|x - 3| \lt 5\) says "within 5 of 3": \(-2 \lt x \lt 8\). And \(|x - 3| \gt 5\) says "further than 5 from 3": \(x \lt -2\) or \(x \gt 8\), two separate pieces.
2. Multiplying by a negative flips the sign. \(-3x \gt 6\) becomes \(x \lt -2\). The trap is multiplying through by something whose sign you do not know, like \(x - 3\). Do not. Move everything to one side instead.
3. Sign charts. A product or quotient changes sign only where a factor is zero or undefined. Mark those points on a number line, test one value in each interval, and record the sign of each factor and of the product in rows, as in the chart below.
And one inequality that is a theorem. For non-negative \(a, b\),
\(\dfrac{a + b}{2} \ge \sqrt{ab}\), with equality exactly when \(a = b\). (AM–GM)
Proof in one line: \((\sqrt a - \sqrt b)^2 \ge 0\), expand to \(a - 2\sqrt{ab} + b \ge 0\), rearrange.
Watch one: two kinds
\(x^2 - x - 6 \lt 0\). Factor: \((x - 3)(x + 2) \lt 0\). Critical points \(-2\) and \(3\). Test \(x = 0\): \((-3)(2) \lt 0\), so the middle interval works; the outer ones give positive products. Solution \(-2 \lt x \lt 3\). (A parabola opening upward is below the axis between its roots.)
\(\dfrac{x - 1}{x + 2} \ge 0\). Critical points \(1\) (allowed, since \(\ge\)) and \(-2\) (never allowed). Test \(-3\): positive, yes. \(0\): negative, no. \(2\): positive, yes. Solution \(x \lt -2\) or \(x \ge 1\). Note that 1 is included (the fraction can be 0) but \(-2\) is not (division by zero).
Your turn
- Solve, saying each answer in words as a distance: \(|x + 4| \lt 3\); \(\;|3 - x| \ge 2\).
- Solve \(x^2 \ge 16\) (many people write only half the answer). Then \(2x^2 - 7x + 3 \gt 0\) by sign chart.
- Solve \(\dfrac{x + 1}{x - 3} \lt 2\). Do not multiply by \(x - 3\); move the 2 across, combine into one fraction, make a sign chart.
- Use AM–GM. (a) A rectangle has perimeter 40 m. Show its area is at most 100 m². (b) Find the smallest value of \(x + \dfrac9x\) for \(x \gt 0\).
Stretch: the triangle inequality
Prove that for all real \(a, b\), \(|a + b| \le |a| + |b|\). Hint: both sides are non-negative, so compare their squares, and use \(ab \le |a||b|\).
Answers
- Within 3 of \(-4\): \(-7 \lt x \lt -1\). \(|3 - x| = |x - 3| \ge 2\), at least 2 from 3: \(x \le 1\) or \(x \ge 5\).
- \((x-4)(x+4) \ge 0\): \(x \le -4\) or \(x \ge 4\). \((2x - 1)(x - 3) \gt 0\): outside the roots, \(x \lt \tfrac12\) or \(x \gt 3\).
- \(\dfrac{x+1}{x-3} - 2 = \dfrac{7 - x}{x - 3} \lt 0\). Critical points 3 (excluded) and 7. Test 0: yes; 5: no; 8: yes. Solution \(x \lt 3\) or \(x \gt 7\). (Multiplying through would silently lose every solution below 3.)
- (a) Sides \(x + y = 20\), so \(\sqrt{xy} \le 10\), \(xy \le 100\), equality for the \(10 \times 10\) square. (b) \(\tfrac{x + 9/x}{2} \ge \sqrt9 = 3\), so the minimum is 6, at \(x = 3\).
Stretch. \((a+b)^2 = a^2 + 2ab + b^2 \le a^2 + 2|a||b| + b^2 = (|a|+|b|)^2\) since \(ab \le |a||b|\). Both bases are non-negative, so \(|a+b| \le |a|+|b|\). Drawn as arrows head to tail, \(a\), \(b\) and \(a+b\) are the three sides of a triangle, hence the name.