Concept understanding — Arithmetic Mean Geometric Mean
The AM–GM Inequality: Why Averages Are Not All Equal
You already know what an average is — add up a few numbers and divide by how many there are. That is the arithmetic mean (AM). For two numbers a and b, it is
AM=2a+b.
Now imagine you want a different kind of "middle" — one that works for ratios, growth rates, or areas. If you multiply the numbers together and take the nth root, you get the geometric mean (GM). For two numbers a and b,
GM=ab.
The geometric mean answers a different question: "If I had a rectangle of sides a and b, what side length would give me the same area in a square?" That square's side is ab.
The Core Idea
Here is the surprising fact: the geometric mean is never larger than the arithmetic mean. They are equal only when all the numbers are identical. For any two non‑negative numbers a and b,
2a+b≥ab,
with equality if and only if a=b.
This is the AM–GM inequality. It is one of the most used inequalities in mathematics — not because it is complicated, but because it is simple and powerful.
Important
AM–GM Inequality (two numbers)
For a,b≥0,
2a+b≥ab,
with equality iff a=b.
Why Should You Believe It?
Take any two non‑negative numbers, say 4 and 9.
Arithmetic mean: (4+9)/2=6.5
Geometric mean: 4×9=36=6
Indeed 6.5>6. Try 1 and 100: AM = 50.5, GM = 10. The gap can be huge.
What about 5 and 5? AM = 5, GM = 5 — equal, because the numbers are equal.
Tip
A quick visual proof: For any a,b≥0, consider (a−b)2≥0. Expanding gives a+b−2ab≥0, so a+b≥2ab, which is exactly the AM–GM inequality.
The General Statement
The same idea works for any number of non‑negative numbers. For n numbers x1,x2,…,xn≥0,
Since both sides of the inequality are sums of positive square roots, comparing their squares (using (x+y)2=x2+2xy+y2) is an easier, equivalent way to establish which side is larger. …
Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.
WBJEE 2022Set math-20221 markMCQ
Q.If (cotα1)(cotα2)…(cotαn)=1, 0<α1,α2,…αn<π/2, then the maximum value of (cosα1)(cosα2)…(cosαn) is given by
(A) 2n/21
(B) 2n1
(C) 2n1
(D) 1
›Reveal solutionSolution
The constraint gives ∏cosαi=∏sinαi; then (∏cosαi)2=∏(21sin2αi)≤(21)n.
Concept.∏cotαi=1 means ∏sinαi∏cosαi=1, so ∏cosαi=∏sinαi. Let P=∏cosαi.
Q.Let a, b, c be real numbers, each greater than 1, such that 32logba+53logcb+25logac=3. If the value of b is 9, then the value of 'a' must be
(A) 381
(B) 227
(C) 18
(D) 27
›Reveal solutionSolution
Product of the three log-terms is 1; AM-GM forces equality, so each weighted term =1.
Let x=logba,y=logcb,z=logac; then xyz=lnblna⋅lnclnb⋅lnalnc=1, and all positive (since a,b,c>1). …