AM-GM-HM inequalities — proofs — Maths notes

WHAT are these three means?

The two-variable case (do this first — it's 80% of the intuition)

The general case: proving $\text{AM}\ge\text{GM}$ for $n$ numbers

There are many proofs. Here is the cleanest self-contained one: Cauchy's forward–backward induction.

Formula

$\text{AM}\ge\text{GM}$ (general), Cauchy's proof Step A — powers of 2 (forward doubling). Base $n=2$ is proved above. Suppose it holds for $n$ . For $2n$ numbers group them into two blocks of $n$ : $\frac{a_1+\cdots+a_{2n}}{2n}=\frac{1}{2}\!\left(\underbrace{\frac{a_1+\cdots+a_n}{n}}_{\ge (a_1\cdots a_n)^{1/n}}+\underbrace{\frac{a_{n+1}+\cdots+a_{2n}}{n}}_{\ge (a_{n+1}\cdots a_{2n})^{1/n}}\right).$ Why? Each block obeys AM–GM (induction hypothesis). Now apply the 2-number AM–GM to the two block-GMs $x=(a_1\cdots a_n)^{1/n}$ and $y=(a_{n+1}\cdots a_{2n})^{1/n}$ : $\frac{x+y}{2}\ge\sqrt{xy}=(a_1\cdots a_{2n})^{1/(2n)}.$ So the statement holds for $2n$ . By induction it holds for all $n=2,4,8,\dots$

Step B — fill the gaps (backward step). Suppose AM–GM holds for $n$ ; show it for $n-1$ . Take $a_1,\dots,a_{n-1}$ and set the $n$ -th number equal to their AM: $a_n = A=\frac{a_1+\cdots+a_{n-1}}{n-1}.$ Why choose this? Because it makes the total sum tidy — the sum of all $n$ numbers becomes $nA$ . Apply AM–GM for $n$ : $\underbrace{\frac{a_1+\cdots+a_{n-1}+A}{n}}_{=A}\ge\big(a_1\cdots a_{n-1}\cdot A\big)^{1/n}.$ Raise to power $n$ : $A^{n}\ge a_1\cdots a_{n-1}\cdot A$ , so $A^{n-1}\ge a_1\cdots a_{n-1}$ , giving $A=\frac{a_1+\cdots+a_{n-1}}{n-1}\ge (a_1\cdots a_{n-1})^{1/(n-1)}.$ Steps A and B together reach every integer $n\ge2$ . Equality throughout iff all $a_i$ equal. ∎

Now $\text{GM}\ge\text{HM}$ in general (free of charge)

Worked examples

Common mistakes (Steel-manned)

Active recall

Recall Feynman: explain to a 12-year-old

Imagine 3 friends have different numbers of candies. There are three "fair-share" ideas. AM = pile all candies, split equally — friendly to big piles. GM = if instead each day you multiplied your candies, this is the fair daily multiplier — it cares about balance. HM = fair only when you think about "candies per second of effort" (rates/reciprocals) — it lets small piles dominate. The magic rule: the "add-average" is always the biggest, the "rate-average" the smallest, and the "multiply-average" is stuck between them. They become equal only when everyone already has the same candies — nothing to fix!

Flashcards

AM formula for

a_1..a_n

\frac{a_1+\cdots+a_n}{n}

GM formula for

a_1..a_n

(a_1a_2\cdots a_n)^{1/n}

HM formula for

a_1..a_n

\dfrac{n}{\sum 1/a_i}

, i.e. reciprocal of AM-of-reciprocals

The master inequality (positives)

\text{AM}\ge\text{GM}\ge\text{HM}

, equality iff all equal

2-number AM≥GM proof idea

(\sqrt a-\sqrt b)^2\ge0 \Rightarrow a+b\ge2\sqrt{ab}

How to get GM≥HM from AM≥GM

apply AM≥GM to reciprocals, then take reciprocals (flip)

Name of the general AM≥GM proof

Cauchy forward–backward (doubling) induction

Forward step of Cauchy proof

prove for

2n

by grouping two blocks of

n

+ 2-term AM–GM

Backward step of Cauchy proof

set

a_n=

AM of the others to descend

n\to n-1

Identity true only for n=2

\text{GM}^2=\text{AM}\cdot\text{HM}

Min of

x+1/x

x>0

2

, at

x=1

(by AM–GM)

Average speed over equal distances u,v

\text{HM}=\frac{2uv}{u+v}

Why AM–GM needs positivity

\sqrt{a_i}

undefined / GM ill-defined for negatives

Connections

Arithmetic Progressions — AM is the middle term of a 3-term AP.
Geometric Progressions — GM is the middle term of a 3-term GP.
Harmonic Progression — HM is the middle term of a 3-term HP.
Cauchy-Schwarz Inequality — another workhorse; AM–GM is a special/limit case.
Jensen's Inequality — AM–GM is Jensen applied to concave $\log$ .
Optimization using inequalities — fixed product ⇒ AM–GM gives extrema.
Weighted Means (Power Mean Inequality) — AM,GM,HM are power means $p=1,0,-1$ .

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho, teen tarah ke "average" hote hain positive numbers ke liye: AM (arithmetic mean, jahan hum jodte hain), GM (geometric mean, jahan hum multiply karke root lete hain), aur HM (harmonic mean, jahan hum reciprocals ka average nikaal ke wapas ulta karte hain). Sabse important rule yaad rakhna: AM ≥ GM ≥ HM, aur equality tabhi aati hai jab saare numbers barabar ho. Matlab jitni zyada spread (difference) hogi numbers mein, utna zyada ye teenon alag ho jayenge.

Proof ka asli jaadu do numbers wale case se aata hai. $(\sqrt a - \sqrt b)^2 \ge 0$ — bas isko expand karo aur AM ≥ GM ready ho jaata hai, kyunki square kabhi negative nahi hota. Phir GM ≥ HM ke liye alag mehnat nahi karni: reciprocals $\frac1a, \frac1b$ pe wahi AM–GM laga do aur ulta kar do (reciprocal lene par inequality flip hoti hai). Yehi trick $n$ numbers ke liye bhi chalti hai.

General proof Cauchy ka forward–backward induction hai: pehle powers of 2 ( $2,4,8,\dots$ ) ke liye prove karo blocks banake, phir ek-ek step neeche utar ke ( $n \to n-1$ ) beech ke saare numbers cover karo. Idea simple hai — seedhi ladder ke rungs sirf $2^k$ pe hain, lekin ek slide hai jo tumhe kisi bhi neeche wale point tak le aati hai.

Kyun matter karta hai? Optimization mein bahut kaam aata hai — jaise $x + \frac1x \ge 2$ , ya average speed (equal distance pe HM lagta hai, AM nahi — isliye 60 aur 40 ka average speed 50 nahi, 48 hai!). Exam mein "minimum/maximum find karo" wale questions mostly AM–GM se ek line mein ud jaate hain. Bas positivity condition mat bhoolna, warna proof toot jaata hai.

AM-GM-HM inequalities — proofs

WHAT are these three means?

The two-variable case (do this first — it's 80% of the intuition)

The general case: proving $\text{AM}\ge\text{GM}$ for $n$ numbers

Now $\text{GM}\ge\text{HM}$ in general (free of charge)

Worked examples

Common mistakes (Steel-manned)

Active recall

Flashcards

Connections

Concept Map

Hinglish (regional understanding)

Go deeper — visual, from zero

Test yourself — Sequences & Series

Connections