Harmonic Progression • Topic 2 of 2

AM-GM-HM Relationship

For any set of positive real numbers, the arithmetic, geometric and harmonic means obey AM ≥ GM ≥ HM, and all three are equal only when every number is the same. This is one of the most exploited results in CAT Algebra and Maxima–Minima. For two positive numbers a and b: AM = (a+b)/2, GM = √(ab), HM = 2ab/(a+b), and they are linked by the beautiful identity GM² = AM × HM — so the GM is itself the geometric mean of the AM and HM. The most common exam use is finding a minimum: by AM ≥ GM, any expression of the form x + 1/x is at least 2 (equality at x=1), and a sum of terms whose product is fixed is minimised when the terms are equal. To apply it, spot a sum whose terms have a constant product, drop to the GM, and read off the bound. Remember the inequality only holds for positive reals — using it on a term that can be negative is the classic trap.

✅ Solved examples

1. Find the minimum value of x + 1/x for x > 0.
By AM ≥ GM: (x + 1/x)/2 ≥ √(x · 1/x) = 1, so x + 1/x ≥ 2, with equality at x = 1. Minimum = 2.
2. For two positive numbers, AM = 10 and GM = 8. Find the HM.
GM² = AM × HM ⇒ 64 = 10 × HM ⇒ HM = 6.4.
3. The product of two positive numbers is 16. Find the least possible value of their sum.
By AM ≥ GM, (a+b)/2 ≥ √(ab) = √16 = 4 ⇒ a+b ≥ 8. Least sum = 8 (at a=b=4).
4. If a, b are positive, show (a+b)(1/a + 1/b) is at least 4 and find when equality holds.
Expand: 2 + a/b + b/a. By AM≥GM, a/b + b/a ≥ 2, so the sum ≥ 4. Equality when a/b = b/a, i.e. a = b.

✏️ Practice — try these, take hints as needed

1. Find the minimum value of 4x + 9/x for x > 0.
Product of the two terms is constant: 4x × 9/x = 36.
AM ≥ GM ⇒ sum ≥ 2√36.
2 × 6.
12 (at x = 3/2)
2. For two positive numbers AM = 25 and HM = 16. Find their GM.
Use GM² = AM × HM.
GM² = 25 × 16 = 400.
Take the square root.
20
3. The sum of two positive numbers is 18. Find the maximum value of their product.
By AM ≥ GM the product is largest when the numbers are equal.
Each = 9.
9 × 9.
81
4. Find the minimum value of (a + b + c)(1/a + 1/b + 1/c) for positive a, b, c.
AM ≥ HM on three numbers gives the bound n².
Here n = 3.
3².
9
5. A positive number and its reciprocal have a sum of 2.5. Without solving the quadratic, can the sum be less than 2?
x + 1/x ≥ 2 always for x > 0.
Equality only at x = 1 where the sum is exactly 2.
So values below 2 are impossible.
No — the minimum possible sum is 2

📝 Topic test — 8 questions

Auto-graded with full solutions; saved to your dashboard. Use the calculator and formula sheet (top-right) any time.

Loading questions…
← HP Basics Take the chapter test →