Harmonic Mean

A car drives from Delhi to Agra (a distance of 200 km) at 80 km/h, and returns the same 200 km at 120 km/h. What is the car's average speed for the round trip?

The tempting answer is \dfrac{80 + 120}{2} = 100 km/h. But that is wrong. The car spent more time at the slower speed, so the slower leg should weigh more heavily.

Total distance: 200 + 200 = 400 km. Time for the first leg: \dfrac{200}{80} = 2.5 hours. Time for the return leg: \dfrac{200}{120} = \dfrac{5}{3} hours. Total time: 2.5 + \dfrac{5}{3} = \dfrac{25}{6} hours.

Average speed: \dfrac{400}{25/6} = \dfrac{400 \times 6}{25} = 96 km/h.

That 96 is not the arithmetic mean of 80 and 120. It is the harmonic mean:

Whenever you travel equal distances at different speeds, the average speed is not the AM — it is the HM. The harmonic mean handles rates naturally because rates are reciprocals of per-unit times, and the HM is built from reciprocals.

The HM of two numbers

Take two positive numbers a and b. Their harmonic mean is

This formula looks different from the AM (\dfrac{a+b}{2}) and the GM (\sqrt{ab}), but it has a clean structure. Rewrite it:

The reciprocal of the HM is the AM of the reciprocals. The HM takes reciprocals, averages them, and takes the reciprocal back. This is exactly the flip-solve-flip pattern from harmonic progressions.

The three numbers \dfrac{1}{a}, \dfrac{1}{H}, \dfrac{1}{b} form an AP. Equivalently, a, H, b form an HP. So the HM of two numbers is the middle term of a three-term HP, just as the AM is the middle term of a three-term AP and the GM is the middle term of a three-term GP.

For the speed example: a = 80, b = 120, and H = \dfrac{2 \times 80 \times 120}{80 + 120} = \dfrac{19200}{200} = 96.

Inserting n harmonic means

Inserting n harmonic means between a and b creates an HP of n + 2 terms starting at a and ending at b. The method follows the same flip-solve-flip pattern as always.

Take a = \dfrac{1}{2} and b = \dfrac{1}{14} with n = 2 means.

Step 1. Flip to reciprocals: \dfrac{1}{a} = 2 and \dfrac{1}{b} = 14. You need an AP of 2 + 2 = 4 terms from 2 to 14.

Step 2. Find the common difference of the AP:

Step 3. The reciprocal AP is 2, 6, 10, 14.

Step 4. Flip back: the HP is \dfrac{1}{2}, \dfrac{1}{6}, \dfrac{1}{10}, \dfrac{1}{14}.

The two inserted harmonic means are \dfrac{1}{6} and \dfrac{1}{10}.

The general recipe: to insert n harmonic means between a and b, form an AP of n + 2 terms from \dfrac{1}{a} to \dfrac{1}{b} with common difference d = \dfrac{1/(b) - 1/(a)}{n + 1}, then take the reciprocal of each AP term.

The inequality chain: HM \leq GM \leq AM

This is the centrepiece. For any two positive numbers a and b:

with equality throughout if and only if a = b.

You already know the right half of this chain from the geometric mean article: \text{GM} \leq \text{AM}. The left half — \text{HM} \leq \text{GM} — completes the picture.

Proof that HM \leq GM

Start from what you already know: \text{GM} \leq \text{AM}. Apply this to the reciprocals \dfrac{1}{a} and \dfrac{1}{b}:

The left side is \dfrac{1}{\sqrt{ab}} = \dfrac{1}{\text{GM}}. The right side is \dfrac{1}{H} = \dfrac{1}{\text{HM}} (since the reciprocal of the HM is the AM of the reciprocals). So:

Since both means are positive, flipping the inequality gives:

Equality holds when \dfrac{1}{a} = \dfrac{1}{b}, i.e., when a = b.

Proof of the full chain (alternative: direct algebraic proof)

Here is a self-contained proof of \text{HM} \leq \text{GM} \leq \text{AM} using only the fact that squares are non-negative.

Part 1: GM \leq AM. From (\sqrt{a} - \sqrt{b})^2 \geq 0:

Part 2: HM \leq GM. From (\sqrt{a} - \sqrt{b})^2 \geq 0:

Multiply both sides of the original inequality \text{GM} \leq \text{AM} in a different way. Since a + b \geq 2\sqrt{ab}:

In both parts, equality requires \sqrt{a} = \sqrt{b}, i.e., a = b.

When equality holds

The condition for \text{HM} = \text{GM} = \text{AM} is a = b. In that case, all three means equal a. As soon as a \neq b, strict inequalities hold: \text{HM} < \text{GM} < \text{AM}.

The more unequal a and b are, the larger the gaps between the three means. For a = 1, b = 100: \text{HM} = \dfrac{200}{101} \approx 1.98, \text{GM} = 10, \text{AM} = 50.5. The HM is pulled strongly toward the smaller number, the AM toward the larger number, and the GM sits between them.

Two worked examples

Common confusions

• "The HM is just another formula to memorise." The formula H = \dfrac{2ab}{a+b} is not arbitrary — it is the reciprocal of the AM of the reciprocals. If you understand reciprocals and the AM, you can re-derive the HM formula from scratch every time.

• "The average of two speeds is always the AM." Only when equal times are spent at each speed. When equal distances are covered at different speeds, the average speed is the HM. The distinction between equal-time and equal-distance averaging is the single most important thing to remember about the HM.

• "The HM can be greater than the AM." Never. For positive numbers, \text{HM} \leq \text{GM} \leq \text{AM} always holds. The HM is the smallest of the three classical means.

• "HM is defined for negative numbers." The standard definition requires positive numbers, because the reciprocals must exist and the means must be comparable. You can formally compute \dfrac{2ab}{a+b} for negative a and b, but the inequality chain breaks.

• "When a = b, the HM is undefined." When a = b, the HM is perfectly defined: H = \dfrac{2a^2}{2a} = a. All three means collapse to the same value a.

Going deeper

The HM for n numbers

For n positive numbers x_1, x_2, \dots, x_n, the harmonic mean is

The denominator is the sum of all reciprocals. The full AM-GM-HM inequality for n numbers reads:

with equality if and only if all x_i are equal.

The identity \text{GM}^2 = \text{AM} \times \text{HM}

For two positive numbers a and b:

This says: the geometric mean of two numbers is also the geometric mean of their arithmetic and harmonic means. The GM sits at the exact multiplicative midpoint of the AM and HM — a fact that is both beautiful and useful.

One consequence: if you know any two of the three means, you can find the third. Given \text{AM} = 10 and \text{GM} = 8, the HM is \dfrac{\text{GM}^2}{\text{AM}} = \dfrac{64}{10} = 6.4. And you can recover a and b themselves: they are the roots of t^2 - 2(\text{AM})t + \text{GM}^2 = 0, i.e., t^2 - 20t + 64 = 0, giving t = 4 or t = 16.

Resistors in parallel revisited

When n equal resistors of resistance R are connected in parallel, the total resistance is \dfrac{R}{n} — which is the reciprocal of the sum of n copies of \dfrac{1}{R}. More generally, for unequal resistances R_1, R_2, \dots, R_n in parallel:

This is not the HM (it would be the HM if there were an extra factor of n in the numerator), but the structure is the same — the total resistance is the reciprocal of the sum of conductances. The HM gives the "average resistance" in the sense that replacing every resistor by a single resistor of value equal to the HM would give the same total resistance as replacing them all by that value.

Where this leads next

The harmonic mean completes the trio of classical means. Together with the AM and GM, it forms an inequality chain that is the starting point for competition-level algebra.

• Harmonic Progression — the sequence whose middle term is the HM of its neighbours, and the foundation for every HM calculation.
• Geometric Mean — the multiplicative mean that sits between the HM and the AM, satisfying \text{GM}^2 = \text{AM} \times \text{HM}.
• Arithmetic Mean — the additive mean that is always at least as large as the other two.
• AM-GM-HM Inequality — the full chain \text{HM} \leq \text{GM} \leq \text{AM} extended to n numbers, with applications to optimisation and competition problems.
• Percentages and Ratios — the arithmetic of ratios and rates that makes the HM the natural mean for "per unit" quantities.