AM ≥ GM ≥ HMJEE Main

The inequality that solves minimum problems — interactive Mathematics simulation for IIT-JEE.

Concept

For positive numbers, the three classical means always stack the same way: AM ≥ GM ≥ HM, with equality only when all numbers are equal. This single inequality is a minimum-finding machine: whenever a product is fixed, the sum is minimised at equality — no calculus needed.

Key formula

\frac{x+y}{2} \ge \sqrt{xy} \ge \frac{2xy}{x+y}, \qquad GM^2 = AM \times HM \;(two\;numbers)

Derivation

AM − GM = $\frac{(\sqrt x - \sqrt y)^2}{2} \ge 0$ — a perfect square, zero only at x = y.

HM = GM²/AM (verify: $\frac{2xy}{x+y} = \frac{xy}{(x+y)/2}$ ), so GM is the geometric mean of AM and HM too — the chain follows.

Classic use: $x + \frac 1x \ge 2\sqrt{x \cdot \frac1x} = 2$ , minimum at x = 1.

Scenarios to explore

AM ≥ GM ≥ HM — The mean inequality chain that solves minimum problems.

Real-world applications

Minimum of sums with fixed product (fencing, box optimisation).
Average speed over equal DISTANCES is the harmonic mean.
Cauchy–Schwarz & power-mean ladders build on it.

JEE exam tips

a + b + c ≥ 3(abc)^{1/3} — n-term version, same equality condition.
For min of ax + b/x: split so the product is constant → min = 2√(ab).
Equality case IS the answer location — always state x = y.

Common mistakes

Applying AM-GM to negative numbers (needs positives).
Claiming a minimum without checking equality is ACHIEVABLE in the domain.
Averaging speeds arithmetically for equal distances (harmonic!).

Exam traps to avoid

x + 1/x ≥ 2 holds for x > 0 but ≤ −2 for x < 0 — sign matters.
Weighted AM-GM exists: exam questions sometimes need unequal splits (x = a + a + b style).