AM-GM Inequality

In the case of two variables, the Arithmetic Mean - Geometric Mean (AM-GM) inequality - ab ≤ (a+b)/2 - is a consequence of a specific case of the Isoperimetric theorem:

Among all rectangles of a given area the square has the least perimeter.

Or, equivalently,

Among all rectangles of a given perimeter the square has the largest area.

This duality of the formulation carries over to the AM-AG inequality.

For positive a, b that satisfy a + b = 2, ab ≤ 1.

Equivalently,

For positive a, b that satisfy ab = 1, a + b ≥ 2.

While trivial, it is often useful while solving problems to keep this interpretation in mind. Here's one example from the 1935 Moscow Mathematical Olympiad:

Find all real solutions of the following system:

x + y = 2
xy - z² = 1.

Solution

Since x + y = 2, xy ≤ 1 so that

1 = xy - z² ≤ 1 - z² < 1,

unless z = 0. To avoid a contradiction (1 < 1), we have to accept z = 0 as the only possibility. But then xy = 1 and x + y = 2, implying x = y = 1.


Related material
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  • Averages, Arithmetic and Harmonic Means
  • Expectation
  • The Size of a Class: Two Viewpoints
  • Averages of divisors of a given integer
  • Family Statistics: an Interactive Gadget
  • Averages in a sequence
  • Arithmetic and Geometric Means
  • Geometric Meaning of the Geometric Mean
  • A Mathematical Rabbit out of an Algebraic Hat
  • The Mean Property of the Mean
  • Harmonic Mean in Geometry
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