3.3.5 · D5Sequences & Series
Question bank — AM-GM-HM inequalities — proofs
Before we start, a quick reminder of the three names so every symbol below is earned. Throughout, == means the number of terms== — how many positive numbers we are averaging:
- AM (Arithmetic Mean) = add them up, divide by (how many) — from Arithmetic Progressions.
- GM (Geometric Mean) = multiply them, take the -th root (the root matching the count ) — from Geometric Progressions.
- HM (Harmonic Mean) = flip each number, average those, flip back — from Harmonic Progression.
- The master chain: , with equality only when all numbers are equal.
True or false — justify
True or false: AM–GM holds for any two real numbers .
False — GM uses , which needs and needs each number for the proof to even parse; with we get but , and is false.
True or false: if all the numbers are equal, then AM = GM = HM exactly.
True — with no spread there is nothing to "punish", so the inequality collapses to equality; algebraically the squared term that drives every proof is exactly .
True or false: making the numbers more spread out (same sum) pushes AM and GM further apart.
True — the gap measures spread; equal numbers give gap , and increasing variance while fixing the sum lowers the product, hence lowers GM while AM stays put.
True or false: for any number of positive reals.
False — take : , but and , so . It is only a two-number miracle where the factors cancel; for no such cancellation exists (some triples may match by coincidence, but the identity does not hold in general).
True or false: the average speed of a there-and-back trip at 60 and 40 is 50.
False — equal distances means times weight the reciprocals, so it is the HM ; AM would be correct only if you spent equal times at each speed.
True or false: HM is just "the AM done to the reciprocals, then flipped back".
True — that is literally its definition: (with the number of terms), i.e. is the reciprocal of the AM of the reciprocals.
True or false: HM ≤ GM is a separate theorem needing its own new idea.
False — it is free: apply the already-proved AM≥GM to the reciprocals , which gives , and flipping positives reverses to .
True or false: Cauchy's forward–backward induction proves AM–GM only for a power of 2.
False — the forward step reaches all powers of 2, then the backward step slides down one at a time from any power of 2 to fill every integer .
True or false: if AM = GM for some positive numbers, they need not all be equal.
False — equality forces every , so all numbers coincide; this equality condition is the whole point of most optimisation answers.
Spot the error
" works for all , so AM≥GM is universal."
The step silently assumed ; for negative inputs the square-roots aren't real, so the whole line is meaningless, not universal.
"I applied AM–GM to and to minimise ."
AM–GM demands positive terms; for , so the inequality does not apply and any "minimum" you extract is invalid.
", so equality happens whenever the product is ."
Equality needs each two-term AM–GM to be tight simultaneously — , , — i.e. ; a coincidental product value isn't enough.
"HM is bigger than AM because HM 'lets small numbers dominate'."
Letting small numbers dominate lowers the average, so HM is the smallest, not biggest — HM ≤ AM always.
"For , AM–GM gives min , so max is unbounded — but it also gives a maximum of ."
AM–GM only bounds from below here; is a minimum, and there is no upper bound since as or .
"I proved AM≥GM for reciprocals, so I've also proven AM≥HM directly."
You proved (i.e. GM≥HM); AM≥HM follows only by chaining AM≥GM≥HM, not from the reciprocal step alone.
"Cauchy's backward step sets to the geometric mean of the others."
It sets to the arithmetic mean of the others, precisely so the total sum becomes and the factors cancel cleanly.
Why questions
Why must all numbers be strictly positive (not just nonzero) for HM?
HM divides by ; a zero makes a reciprocal blow up, and mixed signs can make the denominator vanish or the whole chain of inequalities collapse — positivity keeps every reciprocal finite and same-signed.
Why does the proof of AM≥GM start from squaring specifically?
Squaring that difference manufactures the cross-term , which is exactly the GM we want to isolate — a targeted move, not a lucky one.
Why does taking reciprocals flip an inequality?
For positive quantities, is decreasing, so it reverses order; that reversal is what turns "AM of reciprocals ≥ ..." into "HM ≤ ...".
Why is HM the right average for equal-distance travel but not equal-time?
Equal distances make the times proportional to , so you must average reciprocals (HM); equal times make speeds add linearly, which is AM. See Optimization using inequalities for the pattern.
Why does the two-number identity fail for three or more?
For the sum factor appears in AM and in HM's denominator and cancels; for no such single common factor exists to cancel, so the tidy identity dies.
Why can AM–GM prove optimisation minima but you must always check the equality case?
The inequality only says "at least this big"; the bound is achieved — and hence a true minimum — only when equality holds, which requires all terms equal. See Optimization using inequalities.
Why does the general AM–GM follow from Cauchy-Schwarz-style tricks being unnecessary here?
Because the forward–backward induction is fully self-contained; it never needs Cauchy-Schwarz Inequality or Jensen's Inequality, though those give alternative one-line proofs via convexity.
Edge cases
What is the GM of the numbers ?
It is , since the product is and any root of is ; but then HM is undefined (a reciprocal of ), so the full chain requires strictly positive inputs.
What happens to as two of the numbers become equal?
The gap contributed by that pair shrinks to ; if all numbers coincide the gap vanishes entirely and AM = GM = HM.
For a single number (), what are AM, GM, HM?
All three equal the number itself — with one value there is no spread, so the inequalities are trivial equalities.
As one positive number (others fixed), what happens to GM and HM?
Both : GM because the product , and HM because one reciprocal dominates the denominator — so the smallest number drags the multiplicative and reciprocal averages down hardest.
As one number with the rest fixed, which mean grows without bound and which stays bounded?
AM (the sum blows up) and GM too, but HM stays bounded: as that value its reciprocal , so , meaning HM approaches (but never quite reaches) the finite limit ; this limiting value is at most times the smallest remaining number.
Does AM–GM–HM say anything when exactly one number is huge and the rest tiny?
Yes — the spread is enormous, so the three means are maximally separated: AM chases the huge value, HM clings near the tiny ones, and GM sits (geometrically) between. This is the extreme of "spread punishes".
Can equality hold without ?
No — every equality in the chain forces all numbers equal, so already implies all at once.
For weighted means, does the same AM≥GM≥HM ordering survive?
Yes, with matching weights it generalises to the Weighted Means (Power Mean Inequality): HM, GM, AM are the power means at exponents , and power means increase with the exponent.