AM-GM-HM inequalities — proofs — Hinglish notes

3.3.5 · Maths › Sequences & Series

YE TEEN MEANS HAIN KYA?

Do-variable wala case (pehle ye karo — 80% intuition yahi hai)

General case: $AM \geq GM$ ko $n$ numbers ke liye prove karna

Kai proofs hain. Yahan sabse clean self-contained proof hai: Cauchy's forward–backward induction.

Formula

$AM \geq GM$ (general), Cauchy ka proof Step A — powers of 2 (forward doubling). Base $n = 2$ upar prove ho chuka hai. Maano ye $n$ ke liye hold karta hai. $2 n$ numbers ke liye unhe do $n$ - $n$ ke blocks mein group karo: $\frac{a _{1} + \dots + a _{2 n}}{2 n} = \frac{1}{2} \geq (a_{1} \dots a_{n})^{1/ n} \frac{a _{1} + \dots + a _{n}}{n} + \geq (a_{n + 1} \dots a_{2 n})^{1/ n} \frac{a _{n + 1} + \dots + a _{2 n}}{n} .$ Kyun? Har block AM–GM maanta hai (induction hypothesis). Ab do block-GMs $x = (a_{1} \dots a_{n})^{1/ n}$ aur $y = (a_{n + 1} \dots a_{2 n})^{1/ n}$ par 2-number AM–GM apply karo: $\frac{x + y}{2} \geq x y = (a_{1} \dots a_{2 n})^{1/ (2 n)} .$ To ye statement $2 n$ ke liye bhi hold karta hai. Induction se ye $n = 2, 4, 8, \dots$ sabke liye hold karta hai.

Step B — gaps bharo (backward step). Maano AM–GM $n$ ke liye hold karta hai; dikhao ki ye $n - 1$ ke liye bhi hold karta hai. $a_{1}, \dots, a_{n - 1}$ lo aur $n$ -va number unke AM ke barabar set karo: $a_{n} = A = \frac{a _{1} + \dots + a _{n - 1}}{n - 1} .$ Ye kyun choose kiya? Kyunki isse total sum neat ho jaata hai — saare $n$ numbers ka sum $n A$ ban jaata hai. $n$ ke liye AM–GM apply karo: $= A \frac{a _{1} + \dots + a _{n - 1} + A}{n} \geq (a_{1} \dots a_{n - 1} \cdot A)^{1/ n} .$ Power $n$ se raise karo: $A^{n} \geq a_{1} \dots a_{n - 1} \cdot A$ , isliye $A^{n - 1} \geq a_{1} \dots a_{n - 1}$ , jisse milta hai: $A = \frac{a _{1} + \dots + a _{n - 1}}{n - 1} \geq (a_{1} \dots a_{n - 1})^{1/ (n - 1)} .$ Steps A aur B milke har integer $n \geq 2$ tak pahunchte hain. Equality tab iff saare $a_{i}$ equal hon. ∎

Ab $GM \geq HM$ general case mein (free of charge)

Worked examples

Common mistakes (Steel-manned)

Active recall

Recall Feynman: ek 12-saal ke bacche ko explain karo

Socho 3 doston ke paas alag-alag ginti hain candies ki. "Fair share" ke teen ideas hain. AM = saari candies ikattha karo, barabar baanto — bade piles ke liye friendly. GM = agar har din tum apni candies multiply karte, to ye fair daily multiplier hai — ye balance ki parwah karta hai. HM = tabhi fair hota hai jab tum "candies per second of effort" (rates/reciprocals) ke baare mein socho — ye chote piles ko dominate karne deta hai. Jaadu ka rule: "add-average" hamesha sabse bada hota hai, "rate-average" sabse chota, aur "multiply-average" beech mein fansa rehta hai. Ye equal tab hote hain jab sabke paas pehle se same candies hon — fix karne ko kuch nahi!

Flashcards

a_{1} .. a_{n}

ke liye AM formula

\frac{a _{1} + \dots + a _{n}}{n}

a_{1} .. a_{n}

ke liye GM formula

(a_{1} a_{2} \dots a_{n})^{1/ n}

a_{1} .. a_{n}

ke liye HM formula

\frac{n}{\sum 1/ a _{i}}

, yaani AM-of-reciprocals ka reciprocal

Master inequality (positives ke liye)

AM \geq GM \geq HM

, equality iff saare equal

2-number AM≥GM proof idea

(a - b)^{2} \geq 0 \Rightarrow a + b \geq 2 ab

AM≥GM se GM≥HM kaise nikaalein

reciprocals par AM≥GM apply karo, phir reciprocals lo (flip)

General AM≥GM proof ka naam

Cauchy forward–backward (doubling) induction

Cauchy proof ka forward step

2 n

ke liye prove karo: do

n

n

blocks group karo + 2-term AM–GM

Cauchy proof ka backward step

a_{n} =

baaki ka AM set karo taaki

n \to n - 1

descend ho

Sirf n=2 ke liye sach identity

GM^{2} = AM \cdot HM

x + 1/ x

ka minimum,

x > 0

2

x = 1

par (AM–GM se)

Equal distances u,v par average speed

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

AM–GM ko positivity kyun chahiye

negatives ke liye

a_{i}

undefined / GM ill-defined hota hai

Connections

Arithmetic Progressions — AM, 3-term AP ka middle term hota hai.
Geometric Progressions — GM, 3-term GP ka middle term hota hai.
Harmonic Progression — HM, 3-term HP ka middle term hota hai.
Cauchy-Schwarz Inequality — ek aur workhorse; AM–GM iska special/limit case hai.
Jensen's Inequality — AM–GM, concave $lo g$ par apply Jensen hai.
Optimization using inequalities — fixed product ⇒ AM–GM extrema deta hai.
Weighted Means (Power Mean Inequality) — AM,GM,HM power means hain $p = 1, 0, - 1$ .