AM-GM-HM inequalities — proofs: Worked examples

This page is a drill. The parent note proved the chain $AM \geq GM \geq HM$ . Here we take that proven fact and throw every kind of question at it — so that when an exam or a real problem hands you a case, you have already seen its twin.

Before any symbol appears in an example, here is the vocabulary, in plain words:

Positive real means a number bigger than zero (like $3$ , $0.5$ , $\frac{7}{2}$ ) — never zero, never negative. AM–GM only works here (the parent note's first mistake box shows why).
AM = add them, divide by how many. GM = multiply them, take the root.
HM = flip each number, average the flips, flip back. In symbols, for $n$ positive numbers $a_{1}, \dots, a_{n}$ : $HM = \frac{n}{\frac{1}{a _{1}} + \dots + \frac{1}{a _{n}}}$ . For $n = 2$ this simplifies to $\frac{2 ab}{a + b}$ ; for $n = 3$ all-equal to $t$ it gives $\frac{3}{3/ t} = t$ . Same formula, any $n$ — that's why the 2-term and 3-term versions agree.
"Equality condition" = the exact situation where the two sides are equal, not just $\geq$ . For all three means that situation is: all the numbers are equal.

The scenario matrix

Every question this topic can ask falls into one of these cells. Each worked example below is tagged with the cell it lands in, and together they fill the whole grid.

Cell	Case class	What makes it tricky	Example
C1	Two positives, plain AM≥GM	nothing — the warm-up	Ex 1
C2	Product fixed → minimise a sum	you must spot the constant product	Ex 2
C3	Sum fixed → maximise a product	the mirror image of C2	Ex 3
C4	Three-variable symmetric inequality	apply AM–GM in pieces, multiply	Ex 4
C5	HM word problem (rates / equal distances)	knowing why HM, not AM	Ex 5
C6	Degenerate: all numbers equal	equality case — everything collapses	Ex 6
C7	Limiting / boundary: one number $\to 0$	GM and HM crash toward $0$	Ex 7
C8	The trap: negative inputs	AM–GM breaks — recognise & refuse	Ex 8
C9	Exam twist: weighted / non-obvious split	rig the terms so the product is constant	Ex 9

Ex 1 — Two positives, the raw inequality (Cell C1)

Ex 2 — Fixed product, minimise a sum (Cell C2)

Worked example For

$x > 0$ , find the minimum of $9 x + \frac{4}{x}$ . Forecast: The two pieces $9 x$ and $\frac{4}{x}$ trade off — grow $x$ and the first rises while the second falls. Guess: there is a sweet spot. What's the minimum value?

Check the product is constant. $9 x \cdot \frac{4}{x} = 36$ — the $x$ cancels. Why this step? AM–GM pins a sum from below only when the product is fixed. Constant product $=$ constant GM $=$ a floor on the sum. That is the entire trick of Cell C2.
Apply 2-term AM–GM to $9 x$ and $\frac{4}{x}$ (both positive): $\frac{9 x + \frac{4}{x}}{2} \geq 9 x \cdot \frac{4}{x} = 36 = 6.$ Why this step? We want a lower bound, and AM $\geq$ GM gives exactly that.
Clear the $2$ . $9 x + \frac{4}{x} \geq 12$ . Why this step? Multiply both sides by $2$ to isolate the quantity we care about.
Find where it's reached. Equality iff $9 x = \frac{4}{x}$ , i.e. $x^{2} = \frac{4}{9}$ , so $x = \frac{2}{3}$ . Why this step? AM=GM only when the two terms are equal — that locates the minimiser.

Verify: At $x = \frac{2}{3}$ : $9 \cdot \frac{2}{3} = 6$ and $\frac{4}{2/3} = 6$ , sum $= 12$ . ✓ Minimum value $= 12$ at $x = \frac{2}{3}$ .

Ex 3 — Fixed sum, maximise a product (Cell C3)

Worked example Positive numbers

$a, b, c$ satisfy $a + b + c = 12$ . What is the largest possible value of $ab c$ ? Forecast: With a fixed sum, does the product get big when the numbers are lopsided (like $10, 1, 1$ ) or when they are balanced? Guess before reading.

Write AM–GM for three numbers. $\frac{a + b + c}{3} \geq (ab c)^{1/3}$ . Why this step? Here the sum is fixed, so the left side is a known constant $= \frac{12}{3} = 4$ . AM $\geq$ GM then caps the GM — hence caps the product.
Substitute the sum. $4 \geq (ab c)^{1/3}$ . Why this step? Plugging $a + b + c = 12$ turns the abstract bound into a number.
Cube both sides. $ab c \leq 4^{3} = 64$ . Why this step? Cubing (both sides positive) removes the root and gives the ceiling on $ab c$ .
When is it reached? Equality iff $a = b = c$ . With sum $12$ that forces $a = b = c = 4$ . Why this step? The balanced point is where AM=GM — lopsided splits waste product.

Verify: $4 \cdot 4 \cdot 4 = 64$ and $4 + 4 + 4 = 12$ . Try a lopsided split $10, 1, 1$ : product $= 10 < 64$ . ✓ Maximum $ab c = 64$ .

Ex 4 — Three-variable symmetric inequality (Cell C4)

Worked example For

$a, b, c > 0$ show $\frac{a}{b} + \frac{b}{c} + \frac{c}{a} \geq 3$ . Forecast: Each ratio can be big or small, but they're linked. Guess the smallest possible total.

Name the three positive terms. Let $u = \frac{a}{b}, v = \frac{b}{c}, w = \frac{c}{a}$ . All positive. Why this step? AM–GM wants a clean list of positive numbers; naming them stops us tripping over fractions.
Compute their product. $uv w = \frac{a}{b} \cdot \frac{b}{c} \cdot \frac{c}{a} = 1$ — everything cancels. Why this step? A constant product ( $= 1$ ) is the green light for Cell-C2-style bounding of the sum.
Apply 3-term AM–GM. $\frac{u + v + w}{3} \geq (uv w)^{1/3} = 1^{1/3} = 1$ . Why this step? AM $\geq$ GM turns the fixed product into a floor on the sum.
Clear the $3$ . $u + v + w \geq 3$ , i.e. $\frac{a}{b} + \frac{b}{c} + \frac{c}{a} \geq 3$ . Why this step? Multiply out to recover the original quantity. Equality iff $u = v = w = 1$ , i.e. $a = b = c$ .

Verify: Test $a = 1, b = 2, c = 4$ : $\frac{1}{2} + \frac{1}{2} + 4 = 5 \geq 3$ . ✓ Test $a = b = c = 1$ : $1 + 1 + 1 = 3$ (equality). ✓

Ex 5 — HM word problem, equal distances (Cell C5)

Worked example A drone flies

$30$ km out at $60$ km/h and the same $30$ km back at $40$ km/h. Average speed for the round trip? Forecast: Tempting to say $\frac{60 + 40}{2} = 50$ . Will the true answer be above, below, or exactly $50$ ?

Average speed = total distance ÷ total time. Distance $= 60$ km. Why this step? "Average speed" is defined this way — not as the AM of the two speeds.
Total time. Out: $\frac{30}{60} = 0.5$ h. Back: $\frac{30}{40} = 0.75$ h. Total $= 1.25$ h. Why this step? Equal distances mean the times differ, weighted by the reciprocals of speed — this is where HM comes from.
Divide. Average $= \frac{60}{1.25} = 48$ km/h. Why this step? Executes the definition from step 1.
Recognise it as HM. $HM (60, 40) = \frac{2 \cdot 60 \cdot 40}{60 + 40} = \frac{4800}{100} = 48$ . Why this step? Confirms the shortcut: equal-distance average speed is the harmonic mean.

Verify: $48 < 50 = AM$ , exactly as the chain $HM \leq AM$ promises — the slow leg drags the average down. ✓ Average speed $= 48$ km/h.

Ex 6 — Degenerate case: all numbers equal (Cell C6)

Ex 7 — Limiting behaviour: one input heads to zero (Cell C7)

Worked example Take

$a = 1$ and let $b$ shrink: $b = 1$ , then $0.01$ , then $0.0001$ . Track AM, GM, HM. Forecast: As $b \to 0$ , which means survive and which crash to zero?

$b = 1$ : AM $= 1$ , GM $= 1$ , HM $= 1$ (Cell C6 in disguise).
$b = 0.01$ : AM $= \frac{1.01}{2} = 0.505$ ; GM $= 0.01 = 0.1$ ; HM $= \frac{2 \cdot 1 \cdot 0.01}{1.01} = \frac{0.02}{1.01} \approx 0.0198$ . Why this step? Watch the spread widen the gaps — AM barely moves (it only feels the sum), while GM and HM plummet (they feel the small number hard).
$b = 0.0001$ : AM $\approx 0.50005$ ; GM $= 0.0001 = 0.01$ ; HM $= \frac{0.0002}{1.0001} \approx 0.0002$ . Why this step? Confirms the trend — as $b \to 0$ : $AM \to \frac{1}{2}$ , $GM \to 0$ , $HM \to 0$ (and HM $\approx 2 b$ , the fastest to die).

The figure below plots all three means as $b$ slides from $1$ down toward $0$ (with $a$ pinned at $1$ ). Read it left-to-right as $b$ shrinks: the blue AM curve flattens toward $0.5$ , the green GM curve bends down to $0$ , and the red HM curve dives to $0$ fastest of all — hugging the line $\approx 2 b$ . That single picture is the whole of Cell C7.

Verify: At $b = 0.01$ : check $0.505 \geq 0.1 \geq 0.0198$ — chain holds. ✓ HM $\to 0$ even faster than GM because HM is dominated by the smallest term.

Ex 8 — The forbidden case: negative inputs (Cell C8)

Worked example "Apply AM–GM to

$a = - 4, b = - 9$ ." Is $\frac{a + b}{2} \geq ab$ ? Forecast: The algebra looks the same as always — will it still work?

Compute the AM. $\frac{- 4 + ( - 9 )}{2} = \frac{- 13}{2} = - 6.5$ . Why this step? AM is defined for any reals — this part is harmless.
Compute $ab$ . $ab = (- 4) (- 9) = 36$ , so $36 = 6$ . Why this step? The product of two negatives is positive, so a real square root exists — this is the trap: it looks like a valid GM.
Test the claim. Is $- 6.5 \geq 6$ ? No. The inequality is false. Why this step? Demonstrates the parent's first mistake box: AM–GM needs $a, b > 0$ because $a, b$ must be real, and the proof's $(a - b)^{2}$ collapses for negatives.
The correct verdict. AM–GM does not apply to negative numbers — refuse the question.

Verify: $- 6.5 \geq 6$ is false. ✓ (Falsity confirmed — proving the restriction $a, b > 0$ is essential.)

Ex 9 — Exam twist: rig the split so the product is constant (Cell C9)

Worked example For

$x > 0$ minimise $f (x) = x^{2} + \frac{54}{x}$ . Forecast: Two terms, product $x^{2} \cdot \frac{54}{x} = 54 x$ — not constant. Naive AM–GM would fail. What repair lets it work?

Diagnose why the naive split fails. $x^{2} \cdot \frac{54}{x} = 54 x$ still depends on $x$ . Why this step? AM–GM only pins the sum when the product is a pure constant. So we must break $\frac{54}{x}$ into equal pieces.
Split into three terms so the product is constant: write $f (x) = x^{2} + \frac{27}{x} + \frac{27}{x}$ . Why this step? Now the product is $x^{2} \cdot \frac{27}{x} \cdot \frac{27}{x} = 2 7^{2} = 729$ — free of $x$ . The number of pieces ( $3$ ) and the split ( $27 + 27$ ) are chosen precisely to cancel $x$ .
Apply 3-term AM–GM. $\frac{x ^{2} + \frac{27}{x} + \frac{27}{x}}{3} \geq (729)^{1/3} = 9$ . Why this step? AM $\geq$ GM turns the constant product into a floor.
Clear the $3$ . $f (x) \geq 27$ . Why this step? Multiply out to recover $f$ . Equality holds only when all three terms coincide: $x^{2} = \frac{27}{x} = \frac{27}{x}$ . The last two copies are already equal, so the real condition is $x^{2} = \frac{27}{x}$ , i.e. $x^{3} = 27$ , giving $x = 3$ .

Verify: At $x = 3$ : $3^{2} + \frac{54}{3} = 9 + 18 = 27$ . ✓ Minimum value $= 27$ at $x = 3$ .

Recall

Recall Which cell is which?

Fixed product, minimise sum ::: Cell C2 (Ex 2) — bound sum from below. Fixed sum, maximise product ::: Cell C3 (Ex 3) — bound product from above. Equal-distance average speed ::: Cell C5 (Ex 5) — it's the HM, always $\leq$ AM. Why can't AM–GM take $a = - 4, b = - 9$ ? ::: $a, b$ aren't real; the proof's $(a - b)^{2}$ collapses (Ex 8). Trick when the product isn't constant ::: split a term into equal pieces until it is (Ex 9). Only case with AM=GM=HM ::: all numbers equal (Ex 6).

Related: Optimization using inequalities · Weighted Means (Power Mean Inequality) · Cauchy-Schwarz Inequality · Jensen's Inequality · Harmonic Progression.