☰ Contents · Algebra

Proving simple inequalities

Lessons 17 · 1 lessons · Sh.A. Alimov, A.R. Xalmukhamedov, M.A. Mirzaakhmedov. Algebra: textbook for Grade 9 of general secondary schools. Revised 4th edition. Tashkent: “O‘qituvchi”, 2019
17

Proving simple inequalities

Textbook: pp. 80–83
GoalApply methods of proving simple inequalities: sign of the difference, the AM–GM inequality, and assuming the opposite.
New words
arithmetic mean · o‘rta arifmetikgeometric mean · o‘rta geometrikproof · isbotlashassuming the opposite (contradiction) · teskarisini faraz qilish
Explanation

The simplest way to prove an inequality is the definition: to prove A ≥ B we show A − B ≥ 0, often with a perfect square. For example, a² + b² − 2ab = (a − b)² ≥ 0, so a² + b² ≥ 2ab. For two positive numbers a and b the arithmetic mean is (a + b) : 2 and the geometric mean is √(ab). The inequality (a + b) : 2 ≥ √(ab) holds because the difference is (√a − √b)² : 2 ≥ 0; equality happens only when a = b. With it we can show, for example, that x + 1/x ≥ 2 for x > 0 and find the smallest value of expressions whose product is constant. Another method is assuming the opposite: suppose the inequality fails and reach a contradiction (for example, a square being negative); then the original inequality is true.

Worked examples
a² + b² ≥ 2ab: a² + b² − 2ab = (a − b)² ≥ 0, with equality only for a = b.
For x > 0, x + 4/x ≥ 4: x and 4/x are positive with product 4. (x + 4/x) : 2 ≥ √(x · 4/x) = 2, so x + 4/x ≥ 4; equality when x = 4/x, i.e. x = 2. The smallest value is 4.
Class activity

“Proof chain”: teams prove an inequality such as a² + 9 ≥ 6a by the A − B ≥ 0 method and say when equality holds. Another team looks for mistakes.

Practice
1
Prove x² + 1 ≥ 2x.
2
Find the smallest value of x + 9/x for x > 0.
3
Prove that a/b + b/a ≥ 2 for a, b > 0.
4
When does the arithmetic mean equal the geometric mean, and why?