AM-GM-HM inequalities — proofs: Exercises — Maths deep dive

Intuition How to use this page

Work top-to-bottom. Each problem sits inside a foldable Solution block — try it first with the block closed, then open to check. The levels climb like a staircase: L1 Recognition (spot the mean) → L2 Application (plug the tool) → L3 Analysis (choose which mean and why) → L4 Synthesis (combine several inequalities) → L5 Mastery (full proof, equality tracked).

Two notes on formatting used below. (1) Text wrapped in ==double equals== is highlighted — it just flags a key term worth remembering; read it as ordinary bold. (2) The two-headed arrow $⟺$ and the word "iff" both mean "if and only if" — the statement on the left is true exactly when the statement on the right is true, and vice-versa. So "equality iff all numbers equal" reads: equality happens precisely when the numbers are all equal, and never otherwise.

Everything here rests on the three ideas from the parent note the proofs note: $AM \geq GM \geq HM, equality ⟺ all numbers equal .$ If any symbol feels new, it is defined the first time it is used below.

Level 1 — Recognition

Here you only need to recognise which mean a formula is and read off numbers. No proofs yet.

Recall Solution 1.1

What each symbol means.

$AM$ (Arithmetic Mean) = "add them, split evenly" $= \frac{a + b}{2}$ .
$GM$ (Geometric Mean) = "the number that, multiplied by itself, gives the product" $= ab$ .
$HM$ (Harmonic Mean) = "flip, average, flip back" $= \frac{2}{\frac{1}{a} + \frac{1}{b}} = \frac{2 ab}{a + b}$ .

Plug in $a = 4, b = 9$ : $AM = \frac{4 + 9}{2} = 6.5, GM = 4 \cdot 9 = 36 = 6, HM = \frac{2 \cdot 4 \cdot 9}{13} = \frac{72}{13} \approx 5.538.$ Check the chain: $6.5 \geq 6 \geq 5.538$ . ✓ True, and strict because $4 \neq = 9$ .

Recall Solution 1.2

That formula is the Harmonic Mean $HM$ : it is the reciprocal of the average of the reciprocals. (See Harmonic Progression for where these appear as sequences.) All three means are equal iff every number is the same, $a_{1} = a_{2} = \dots = a_{n}$ . Why? Take just two numbers to see the mechanism. The proof of $AM \geq GM$ starts from a square that can never be negative, $(a - b)^{2} \geq 0$ . A square appears precisely because a real number squared is always $\geq 0$ — that is what forces the inequality. Expanding gives $a - 2 ab + b \geq 0$ , i.e. $\frac{a + b}{2} \geq ab$ . The only way the gap closes to an equality is when that square is exactly zero, $(a - b)^{2} = 0$ , which means $a = b$ , i.e. $a = b$ . Any difference between the numbers makes the square strictly positive and pushes AM strictly above GM. The same "spread ⇒ strict gap" logic runs through all three means, so equality everywhere demands zero spread — all numbers equal.

Level 2 — Application

Now apply the standard 2-term tool $\frac{x + y}{2} \geq x y$ directly.

Recall Solution 2.1

Why AM–GM here? The two positive pieces $x$ and $\frac{9}{x}$ have a constant product $x \cdot \frac{9}{x} = 9$ . Whenever a sum has fixed product, AM–GM pins the smallest possible sum. $\frac{x + \frac{9}{x}}{2} \geq x \cdot \frac{9}{x} = 9 = 3 ⟹ x + \frac{9}{x} \geq 6.$ Equality needs the two pieces equal: $x = \frac{9}{x} \Rightarrow x^{2} = 9 \Rightarrow x = 3$ (positive root). Minimum value $= 6$ at $x = 3$ . (Related idea: Optimization using inequalities.)

Recall Solution 2.2

The two pieces $\frac{a}{b}$ and $\frac{b}{a}$ are positive with product $\frac{a}{b} \cdot \frac{b}{a} = 1$ . $\frac{\frac{a}{b} + \frac{b}{a}}{2} \geq \frac{a}{b} \cdot \frac{b}{a} = 1 ⟹ \frac{a}{b} + \frac{b}{a} \geq 2.$ Equality when $\frac{a}{b} = \frac{b}{a}$ , i.e. $a^{2} = b^{2}$ , i.e. $a = b$ (both positive). ✓

Level 3 — Analysis

Here you must decide which mean the situation demands, and justify it.

The figure below draws the trip as a two-lane road: A and B are the endpoints. The blue arrow (top) is the outbound leg at $60$ km/h taking time $t_{1} = d /60$ ; the pink arrow (bottom) is the return leg at $40$ km/h taking the longer time $t_{2} = d /40$ . The key thing to see: both legs cover the same distance $d$ , yet the slow leg's arrow corresponds to a larger time, so it occupies more of the total clock — which is exactly why the average is pulled toward the slow speed (down to $48$ , not $50$ ).

Figure legend. Two lanes of a road join point $A$ (left) to point $B$ (right), a fixed distance $d$ apart. Top (blue) arrow: the outbound trip at $60$ km/h, clock time $t_{1} = d /60$ . Bottom (pink) arrow: the return trip at $40$ km/h, clock time $t_{2} = d /40$ , which is longer because the speed is smaller. The yellow caption records the final average speed, the harmonic mean $2 d / (t_{1} + t_{2}) = 48$ km/h. Read it as: same distance both ways, but the slow leg claims more of the total time, so it dominates the average.

Recall Solution 3.1

Average speed = total distance ÷ total time. As the figure shows, each leg is length $d$ , but the two legs take different times. $t_{1} = \frac{d}{60}, t_{2} = \frac{d}{40}, avg = \frac{2 d}{t _{1} + t _{2}} = \frac{2 d}{\frac{d}{60} + \frac{d}{40}} = \frac{2}{\frac{1}{60} + \frac{1}{40}} .$ The $d$ cancels, leaving exactly the Harmonic Mean of $60$ and $40$ : $HM = \frac{2 \cdot 60 \cdot 40}{60 + 40} = \frac{4800}{100} = 48 km/h .$ Why HM and not AM? Equal distances means the time spent on each leg is proportional to the reciprocal of its speed — and HM is precisely the mean that averages reciprocals. Because $HM \leq AM$ , the answer $48$ is below the naive $\frac{60 + 40}{2} = 50$ : the slow leg eats more clock time, so it drags the average down.

Recall Solution 3.2

By definition $HM (R_{1}, R_{2}) = \frac{2}{\frac{1}{R _{1}} + \frac{1}{R _{2}}}$ , so $R_{∥} = \frac{1}{\frac{1}{R _{1}} + \frac{1}{R _{2}}} = \frac{1}{2} \cdot \frac{2}{\frac{1}{R _{1}} + \frac{1}{R _{2}}} = \frac{1}{2} HM (R_{1}, R_{2}) .$ Since $HM \leq GM \leq AM$ , we get the clean upper bound $R_{∥} = \frac{1}{2} HM \leq \frac{1}{2} GM = \frac{1}{2} R_{1} R_{2} .$ With $R_{1} = R_{2} = R$ : $R_{∥} = \frac{1}{2} R$ , which is indeed less than each resistor — the well-known "parallel halving".

Level 4 — Synthesis

Combine several AM–GM applications into one inequality.

Recall Solution 4.1

Each factor is a sum of two positives, the ideal shape for 2-term AM–GM: $a + b \geq 2 ab, b + c \geq 2 b c, c + a \geq 2 c a .$ Why may we multiply these three inequalities? Multiplying an inequality $X \geq Y$ by a positive number keeps its direction (a positive scale factor never flips order). Here every left side and every right side is positive, so multiplying $a + b \geq 2 ab$ by the positive quantity $b + c$ gives $(a + b) (b + c) \geq 2 ab (b + c)$ , and then by the positive $c + a$ again — chaining these positive multiplications preserves $\geq$ throughout. Hence we may multiply all three left sides and all three right sides: $(a + b) (b + c) (c + a) \geq 8 ab \cdot b c \cdot c a = 8 a^{2} b^{2} c^{2} = 8 ab c .$ Equality needs all three sub-equalities at once: $a = b$ , $b = c$ , $c = a$ , i.e. $a = b = c$ . ∎

Recall Solution 4.2

The tool: general $n$ -term AM–GM. For any positive numbers $x_{1}, \dots, x_{n}$ , $AM \frac{x _{1} + x _{2} + \dots + x _{n}}{n} \geq GM (x_{1} x_{2} \dots x_{n})^{1/ n}, equality iff all x_{i} equal.$ In words: the "add-and-split" average is never below the "multiply-and-take-root" average. (This is exactly the statement proved by Cauchy's forward–backward induction in the parent note; here $n = 3$ .) Why this tool? Our three terms $\frac{a}{b}, \frac{b}{c}, \frac{c}{a}$ are positive and their product is a constant, $\frac{a}{b} \cdot \frac{b}{c} \cdot \frac{c}{a} = 1$ , so GM is fixed — precisely the situation where AM–GM pins a lower bound on the sum. Apply the $n = 3$ case: $\frac{\frac{a}{b} + \frac{b}{c} + \frac{c}{a}}{3} \geq (\frac{a}{b} \cdot \frac{b}{c} \cdot \frac{c}{a})^{1/3} = 1^{1/3} = 1.$ Multiply by $3$ : $\frac{a}{b} + \frac{b}{c} + \frac{c}{a} \geq 3$ . Equality when all three equal, giving $a = b = c$ . ∎

Recall Solution 4.3

Idea: connect AM and HM of $a, b, c$ . The HM–AM inequality says $HM \leq AM$ , i.e. $\frac{3}{\frac{1}{a} + \frac{1}{b} + \frac{1}{c}} \leq \frac{a + b + c}{3} = \frac{1}{3} .$ Take reciprocals. Because both sides are positive, reciprocating reverses the inequality direction (a bigger positive number has a smaller reciprocal), so $\leq$ becomes $\geq$ : $\frac{\frac{1}{a} + \frac{1}{b} + \frac{1}{c}}{3} \geq 3 ⟹ \frac{1}{a} + \frac{1}{b} + \frac{1}{c} \geq 9.$ Equality at $a = b = c = \frac{1}{3}$ . ∎ (This is the classic bridge; compare Cauchy-Schwarz Inequality, which proves the same via $(a + b + c) (\frac{1}{a} + \frac{1}{b} + \frac{1}{c}) \geq 9$ .)

Level 5 — Mastery

Full proofs with equality tracked; you build the machinery yourself.

Recall Solution 5.1

Step 1 — get $n = 4$ from $n = 2$ (forward doubling). For four positives group into two pairs: $\frac{p + q + r + s}{4} = \frac{1}{2} (\frac{p + q}{2} + \frac{r + s}{2}) \geq \frac{1}{2} (pq + r s) \geq pq r s = (pq r s)^{1/4} .$ Track each 2-term step and its own equality check. The first $\geq$ used 2-term AM–GM on $(p, q)$ and on $(r, s)$ ; its equalities need $p = q$ and $r = s$ . The second $\geq$ used 2-term AM–GM on the two positive numbers $pq$ and $r s$ ; its equality needs $pq = r s$ . Note both $pq > 0$ and $r s > 0$ (products of positives), so the final quantity $pq r s$ is a genuine positive real and raising to the power $1/4$ is legal. So AM–GM holds for $n = 4$ , with equality iff $p = q$ , $r = s$ and $pq = r s$ all hold — i.e. $p = q = r = s$ .

Step 2 — descend from $4$ to $3$ (backward step). Take $a, b, c$ and set the 4th number equal to their average $A = \frac{a + b + c}{3}$ . Then the four numbers sum to $3 A + A = 4 A$ , so their AM is $A$ . Apply the $n = 4$ result: $A = \frac{a + b + c + A}{4} \geq (ab c \cdot A)^{1/4} .$ Raise both sides to the 4th power (both sides positive, so this preserves $\geq$ ): $A^{4} \geq ab c \cdot A$ , hence $A^{3} \geq ab c$ , hence $A = \frac{a + b + c}{3} \geq (ab c)^{1/3} .$ Equality requires equality at every 2-term step, which forces $a = b = c$ . ∎

Recall Solution 5.2

Left inequality $AM \geq GM$ : proved in Problem 5.1. Right inequality $GM \geq HM$ : apply the just-proved AM–GM to the reciprocals $\frac{1}{a}, \frac{1}{b}, \frac{1}{c}$ : $\frac{\frac{1}{a} + \frac{1}{b} + \frac{1}{c}}{3} \geq (\frac{1}{a} \cdot \frac{1}{b} \cdot \frac{1}{c})^{1/3} = \frac{1}{( ab c ) ^{1/3}} = \frac{1}{GM} .$ The left side is $\frac{1}{HM}$ . So $\frac{1}{HM} \geq \frac{1}{GM}$ ; both positive, so flip: $GM \geq HM$ . Chaining, $AM \geq GM \geq HM$ , equality throughout iff $a = b = c$ . ∎ (For the general weighted picture see Weighted Means (Power Mean Inequality) and Jensen's Inequality.)

Recall Solution 5.3

Compute the three means of $1, 2, 4$ . $AM = \frac{1 + 2 + 4}{3} = \frac{7}{3}, GM = (1 \cdot 2 \cdot 4)^{1/3} = 8^{1/3} = 2,$ $HM = \frac{3}{\frac{1}{1} + \frac{1}{2} + \frac{1}{4}} = \frac{3}{\frac{7}{4}} = \frac{12}{7} .$ Form both sides of the candidate identity. $AM \cdot HM = \frac{7}{3} \cdot \frac{12}{7} = \frac{12}{3} = 4, GM^{2} = 2^{2} = 4.$ For these numbers the two sides happen to agree, $4 = 4$ — which is tempting to read as "the identity survives". But one match is not a proof. Try a second triple $a = 1, b = 1, c = 8$ : $AM = \frac{1 + 1 + 8}{3} = \frac{10}{3}, GM = 8^{1/3} = 2, HM = \frac{3}{1 + 1 + \frac{1}{8}} = \frac{3}{\frac{17}{8}} = \frac{24}{17} .$ Now $AM \cdot HM = \frac{10}{3} \cdot \frac{24}{17} = \frac{240}{51} = \frac{80}{17} \approx 4.71$ , while $GM^{2} = 2^{2} = 4$ . Since $4.71 \neq = 4$ , the identity fails. Lesson. $GM^{2} = AM \cdot HM$ is a genuine two-variable fact — it rests on the $(a + b)$ factor cancelling between $AM = \frac{a + b}{2}$ and $HM = \frac{2 ab}{a + b}$ . No such cancellation exists for three or more numbers, so the identity need not generalise: the first triple's agreement was a numerical coincidence, and the second triple is the counterexample that settles it. ∎

Recall One-screen summary (cover and recall)

AM of 4 and 9 ::: $6.5$ HM of 60 and 40 ::: $48$ Min of $x + \frac{9}{x}$ for $x > 0$ ::: $6$ at $x = 3$ Why average speed uses HM ::: equal distances ⇒ time ∝ 1/speed ⇒ reciprocal averaging When AM=GM=HM ::: all numbers equal Does $GM^{2} = AM \cdot HM$ hold for $n = 3$ ? ::: no — two-variable miracle only